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Integral Calculus

86233 $\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x =$

1

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c

2

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + c

3

\frac{1}{3} \tan^{- 1} (\frac{x + \frac{1}{x}}{3}) + c

4

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{x}}{\sqrt{3}}) + c

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) dx = dt$
$∴ I = \int \frac{1}{t^{2} + 3} dt$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + c$
$[∵ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$

MHT CET-2020

Integral Calculus

86234 $\int \frac{d x}{(x + 2) \sqrt{x + 1}} =$

1

2 \tan^{- 1} (\sqrt{x + 1}) + c

2

\tan^{- 1} (\sqrt{x + 1}) + c

3

2 \tan^{- 1} (\sqrt{x + 2}) + c

4

\tan^{- 1} (\sqrt{x + 2}) + c

Explanation:

(A) : $I = \int \frac{d x}{(x + 2) \sqrt{x + 1}}$
Let, $\sqrt{x + 1} = t$
$x + 1 = t^{2} \Rightarrow dx = 2 t \cdot dt$
$∴ I = \int \frac{2 t \cdot dt}{(t^{2} + 1) \cdot t} = 2 \int \frac{dt}{t^{2} + 1}$
$= 2 \tan^{- 1} (t) + c = 2 \tan^{- 1} (\sqrt{x + 1}) + c$

MHT CET-2020

Integral Calculus

86235 $\int x^{3} \cdot e^{x^{2}} dx =$

1

\frac{1}{2} {ex}^{x^{2}} (x^{2} + 1) + c

2

\frac{1}{2} e^{x^{2}} (x^{2} - 1) + c

3

\frac{1}{2} e^{x} (x^{2} + 1) + c

4

\frac{1}{2} e^{x} (x^{2} - 1) + c

Explanation:

(B) : $I = \int x^{3} e^{x^{2}} d x$
Let, $x^{2} = t$
$2 x d x = d t \Rightarrow x \cdot d x = \frac{d t}{2}$
$∴ I = \int t \cdot e^{t} \cdot \frac{d t}{2}$
$= \frac{1}{2} [e^{t} t - e^{t} + c] = \frac{1}{2} [x^{2} e^{x^{2}} - e^{x^{2}} + c]$
$= \frac{1}{2} e^{x^{2}} (x^{2} - 1) + c$

MHT CET-2020

Integral Calculus

86236 $\int \frac{d x}{\sqrt{(x - 1) (x - 2)}} =$

1

\log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

2

\log | (x - \frac{3}{2}) - \sqrt{x^{2} - 3 x + 2} | + c

3

\log | (x + \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

4

\log | (x - 1) + \sqrt{x^{2} - 3 x + 2} | + c

Explanation:

(A) : $I = \int \frac{d x}{\sqrt{(x - 1) (x - 2)}}$
$= \int \frac{d x}{\sqrt{x^{2} - 3 x + 2}}$
$= \int \frac{d x}{\sqrt{x^{2} - 2 \times \frac{3}{2} x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2}}$
$= \int \frac{dx}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}}$
$= \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}} | + c$
$[∵ \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]$
$= \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c$

MHT CET-2020

Integral Calculus

86237 $\int e^{x} \cdot \sec x (1 + \tan x) d x =$

1

e^{x} \sec x + c

2

e^{x} \tan x + c

3

e^{x} \cos ec x + c

4

e^{x} \cot x + c

Explanation:

(A) : $I = \int e^{x} \cdot \sec x (1 + \tan x) d x$
$= \int e^{x} (\sec x + \sec x \cdot \tan x) d x$
Let, $e^{x} . \sec x = t$
$(e^{x} \cdot \sec x + e^{x} \cdot \sec x \cdot \tan x) d x = d t$
$e^{x} (\sec x + \sec x \cdot \tan x) dx = dt$
$∴ I = \int dt = t + c$
$I = e^{x} \cdot \sec x + c$

MHT CET-2020

Integral Calculus

86233 $\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x =$

1

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c

2

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + c

3

\frac{1}{3} \tan^{- 1} (\frac{x + \frac{1}{x}}{3}) + c

4

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{x}}{\sqrt{3}}) + c

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) dx = dt$
$∴ I = \int \frac{1}{t^{2} + 3} dt$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + c$
$[∵ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$

MHT CET-2020

Integral Calculus

86234 $\int \frac{d x}{(x + 2) \sqrt{x + 1}} =$

1

2 \tan^{- 1} (\sqrt{x + 1}) + c

2

\tan^{- 1} (\sqrt{x + 1}) + c

3

2 \tan^{- 1} (\sqrt{x + 2}) + c

4

\tan^{- 1} (\sqrt{x + 2}) + c

Explanation:

(A) : $I = \int \frac{d x}{(x + 2) \sqrt{x + 1}}$
Let, $\sqrt{x + 1} = t$
$x + 1 = t^{2} \Rightarrow dx = 2 t \cdot dt$
$∴ I = \int \frac{2 t \cdot dt}{(t^{2} + 1) \cdot t} = 2 \int \frac{dt}{t^{2} + 1}$
$= 2 \tan^{- 1} (t) + c = 2 \tan^{- 1} (\sqrt{x + 1}) + c$

MHT CET-2020

Integral Calculus

86235 $\int x^{3} \cdot e^{x^{2}} dx =$

1

\frac{1}{2} {ex}^{x^{2}} (x^{2} + 1) + c

2

\frac{1}{2} e^{x^{2}} (x^{2} - 1) + c

3

\frac{1}{2} e^{x} (x^{2} + 1) + c

4

\frac{1}{2} e^{x} (x^{2} - 1) + c

Explanation:

(B) : $I = \int x^{3} e^{x^{2}} d x$
Let, $x^{2} = t$
$2 x d x = d t \Rightarrow x \cdot d x = \frac{d t}{2}$
$∴ I = \int t \cdot e^{t} \cdot \frac{d t}{2}$
$= \frac{1}{2} [e^{t} t - e^{t} + c] = \frac{1}{2} [x^{2} e^{x^{2}} - e^{x^{2}} + c]$
$= \frac{1}{2} e^{x^{2}} (x^{2} - 1) + c$

MHT CET-2020

Integral Calculus

86236 $\int \frac{d x}{\sqrt{(x - 1) (x - 2)}} =$

1

\log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

2

\log | (x - \frac{3}{2}) - \sqrt{x^{2} - 3 x + 2} | + c

3

\log | (x + \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

4

\log | (x - 1) + \sqrt{x^{2} - 3 x + 2} | + c

Explanation:

(A) : $I = \int \frac{d x}{\sqrt{(x - 1) (x - 2)}}$
$= \int \frac{d x}{\sqrt{x^{2} - 3 x + 2}}$
$= \int \frac{d x}{\sqrt{x^{2} - 2 \times \frac{3}{2} x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2}}$
$= \int \frac{dx}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}}$
$= \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}} | + c$
$[∵ \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]$
$= \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c$

MHT CET-2020

Integral Calculus

86237 $\int e^{x} \cdot \sec x (1 + \tan x) d x =$

1

e^{x} \sec x + c

2

e^{x} \tan x + c

3

e^{x} \cos ec x + c

4

e^{x} \cot x + c

Explanation:

(A) : $I = \int e^{x} \cdot \sec x (1 + \tan x) d x$
$= \int e^{x} (\sec x + \sec x \cdot \tan x) d x$
Let, $e^{x} . \sec x = t$
$(e^{x} \cdot \sec x + e^{x} \cdot \sec x \cdot \tan x) d x = d t$
$e^{x} (\sec x + \sec x \cdot \tan x) dx = dt$
$∴ I = \int dt = t + c$
$I = e^{x} \cdot \sec x + c$

MHT CET-2020

Integral Calculus

86233 $\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x =$

1

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c

2

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + c

3

\frac{1}{3} \tan^{- 1} (\frac{x + \frac{1}{x}}{3}) + c

4

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{x}}{\sqrt{3}}) + c

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) dx = dt$
$∴ I = \int \frac{1}{t^{2} + 3} dt$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + c$
$[∵ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$

MHT CET-2020

Integral Calculus

86234 $\int \frac{d x}{(x + 2) \sqrt{x + 1}} =$

1

2 \tan^{- 1} (\sqrt{x + 1}) + c

2

\tan^{- 1} (\sqrt{x + 1}) + c

3

2 \tan^{- 1} (\sqrt{x + 2}) + c

4

\tan^{- 1} (\sqrt{x + 2}) + c

Explanation:

(A) : $I = \int \frac{d x}{(x + 2) \sqrt{x + 1}}$
Let, $\sqrt{x + 1} = t$
$x + 1 = t^{2} \Rightarrow dx = 2 t \cdot dt$
$∴ I = \int \frac{2 t \cdot dt}{(t^{2} + 1) \cdot t} = 2 \int \frac{dt}{t^{2} + 1}$
$= 2 \tan^{- 1} (t) + c = 2 \tan^{- 1} (\sqrt{x + 1}) + c$

MHT CET-2020

Integral Calculus

86235 $\int x^{3} \cdot e^{x^{2}} dx =$

1

\frac{1}{2} {ex}^{x^{2}} (x^{2} + 1) + c

2

\frac{1}{2} e^{x^{2}} (x^{2} - 1) + c

3

\frac{1}{2} e^{x} (x^{2} + 1) + c

4

\frac{1}{2} e^{x} (x^{2} - 1) + c

Explanation:

(B) : $I = \int x^{3} e^{x^{2}} d x$
Let, $x^{2} = t$
$2 x d x = d t \Rightarrow x \cdot d x = \frac{d t}{2}$
$∴ I = \int t \cdot e^{t} \cdot \frac{d t}{2}$
$= \frac{1}{2} [e^{t} t - e^{t} + c] = \frac{1}{2} [x^{2} e^{x^{2}} - e^{x^{2}} + c]$
$= \frac{1}{2} e^{x^{2}} (x^{2} - 1) + c$

MHT CET-2020

Integral Calculus

86236 $\int \frac{d x}{\sqrt{(x - 1) (x - 2)}} =$

1

\log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

2

\log | (x - \frac{3}{2}) - \sqrt{x^{2} - 3 x + 2} | + c

3

\log | (x + \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

4

\log | (x - 1) + \sqrt{x^{2} - 3 x + 2} | + c

Explanation:

(A) : $I = \int \frac{d x}{\sqrt{(x - 1) (x - 2)}}$
$= \int \frac{d x}{\sqrt{x^{2} - 3 x + 2}}$
$= \int \frac{d x}{\sqrt{x^{2} - 2 \times \frac{3}{2} x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2}}$
$= \int \frac{dx}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}}$
$= \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}} | + c$
$[∵ \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]$
$= \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c$

MHT CET-2020

Integral Calculus

86237 $\int e^{x} \cdot \sec x (1 + \tan x) d x =$

1

e^{x} \sec x + c

2

e^{x} \tan x + c

3

e^{x} \cos ec x + c

4

e^{x} \cot x + c

Explanation:

(A) : $I = \int e^{x} \cdot \sec x (1 + \tan x) d x$
$= \int e^{x} (\sec x + \sec x \cdot \tan x) d x$
Let, $e^{x} . \sec x = t$
$(e^{x} \cdot \sec x + e^{x} \cdot \sec x \cdot \tan x) d x = d t$
$e^{x} (\sec x + \sec x \cdot \tan x) dx = dt$
$∴ I = \int dt = t + c$
$I = e^{x} \cdot \sec x + c$

MHT CET-2020

Integral Calculus

86233 $\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x =$

1

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c

2

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + c

3

\frac{1}{3} \tan^{- 1} (\frac{x + \frac{1}{x}}{3}) + c

4

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{x}}{\sqrt{3}}) + c

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) dx = dt$
$∴ I = \int \frac{1}{t^{2} + 3} dt$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + c$
$[∵ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$

MHT CET-2020

Integral Calculus

86234 $\int \frac{d x}{(x + 2) \sqrt{x + 1}} =$

1

2 \tan^{- 1} (\sqrt{x + 1}) + c

2

\tan^{- 1} (\sqrt{x + 1}) + c

3

2 \tan^{- 1} (\sqrt{x + 2}) + c

4

\tan^{- 1} (\sqrt{x + 2}) + c

Explanation:

(A) : $I = \int \frac{d x}{(x + 2) \sqrt{x + 1}}$
Let, $\sqrt{x + 1} = t$
$x + 1 = t^{2} \Rightarrow dx = 2 t \cdot dt$
$∴ I = \int \frac{2 t \cdot dt}{(t^{2} + 1) \cdot t} = 2 \int \frac{dt}{t^{2} + 1}$
$= 2 \tan^{- 1} (t) + c = 2 \tan^{- 1} (\sqrt{x + 1}) + c$

MHT CET-2020

Integral Calculus

86235 $\int x^{3} \cdot e^{x^{2}} dx =$

1

\frac{1}{2} {ex}^{x^{2}} (x^{2} + 1) + c

2

\frac{1}{2} e^{x^{2}} (x^{2} - 1) + c

3

\frac{1}{2} e^{x} (x^{2} + 1) + c

4

\frac{1}{2} e^{x} (x^{2} - 1) + c

Explanation:

(B) : $I = \int x^{3} e^{x^{2}} d x$
Let, $x^{2} = t$
$2 x d x = d t \Rightarrow x \cdot d x = \frac{d t}{2}$
$∴ I = \int t \cdot e^{t} \cdot \frac{d t}{2}$
$= \frac{1}{2} [e^{t} t - e^{t} + c] = \frac{1}{2} [x^{2} e^{x^{2}} - e^{x^{2}} + c]$
$= \frac{1}{2} e^{x^{2}} (x^{2} - 1) + c$

MHT CET-2020

Integral Calculus

86236 $\int \frac{d x}{\sqrt{(x - 1) (x - 2)}} =$

1

\log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

2

\log | (x - \frac{3}{2}) - \sqrt{x^{2} - 3 x + 2} | + c

3

\log | (x + \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

4

\log | (x - 1) + \sqrt{x^{2} - 3 x + 2} | + c

Explanation:

(A) : $I = \int \frac{d x}{\sqrt{(x - 1) (x - 2)}}$
$= \int \frac{d x}{\sqrt{x^{2} - 3 x + 2}}$
$= \int \frac{d x}{\sqrt{x^{2} - 2 \times \frac{3}{2} x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2}}$
$= \int \frac{dx}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}}$
$= \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}} | + c$
$[∵ \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]$
$= \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c$

MHT CET-2020

Integral Calculus

86237 $\int e^{x} \cdot \sec x (1 + \tan x) d x =$

1

e^{x} \sec x + c

2

e^{x} \tan x + c

3

e^{x} \cos ec x + c

4

e^{x} \cot x + c

Explanation:

(A) : $I = \int e^{x} \cdot \sec x (1 + \tan x) d x$
$= \int e^{x} (\sec x + \sec x \cdot \tan x) d x$
Let, $e^{x} . \sec x = t$
$(e^{x} \cdot \sec x + e^{x} \cdot \sec x \cdot \tan x) d x = d t$
$e^{x} (\sec x + \sec x \cdot \tan x) dx = dt$
$∴ I = \int dt = t + c$
$I = e^{x} \cdot \sec x + c$

MHT CET-2020

Integral Calculus

86233 $\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x =$

1

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c

2

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + c

3

\frac{1}{3} \tan^{- 1} (\frac{x + \frac{1}{x}}{3}) + c

4

\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{x}}{\sqrt{3}}) + c

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) dx = dt$
$∴ I = \int \frac{1}{t^{2} + 3} dt$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + c$
$[∵ \int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c]$
$= \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$

MHT CET-2020

Integral Calculus

86234 $\int \frac{d x}{(x + 2) \sqrt{x + 1}} =$

1

2 \tan^{- 1} (\sqrt{x + 1}) + c

2

\tan^{- 1} (\sqrt{x + 1}) + c

3

2 \tan^{- 1} (\sqrt{x + 2}) + c

4

\tan^{- 1} (\sqrt{x + 2}) + c

Explanation:

(A) : $I = \int \frac{d x}{(x + 2) \sqrt{x + 1}}$
Let, $\sqrt{x + 1} = t$
$x + 1 = t^{2} \Rightarrow dx = 2 t \cdot dt$
$∴ I = \int \frac{2 t \cdot dt}{(t^{2} + 1) \cdot t} = 2 \int \frac{dt}{t^{2} + 1}$
$= 2 \tan^{- 1} (t) + c = 2 \tan^{- 1} (\sqrt{x + 1}) + c$

MHT CET-2020

Integral Calculus

86235 $\int x^{3} \cdot e^{x^{2}} dx =$

1

\frac{1}{2} {ex}^{x^{2}} (x^{2} + 1) + c

2

\frac{1}{2} e^{x^{2}} (x^{2} - 1) + c

3

\frac{1}{2} e^{x} (x^{2} + 1) + c

4

\frac{1}{2} e^{x} (x^{2} - 1) + c

Explanation:

(B) : $I = \int x^{3} e^{x^{2}} d x$
Let, $x^{2} = t$
$2 x d x = d t \Rightarrow x \cdot d x = \frac{d t}{2}$
$∴ I = \int t \cdot e^{t} \cdot \frac{d t}{2}$
$= \frac{1}{2} [e^{t} t - e^{t} + c] = \frac{1}{2} [x^{2} e^{x^{2}} - e^{x^{2}} + c]$
$= \frac{1}{2} e^{x^{2}} (x^{2} - 1) + c$

MHT CET-2020

Integral Calculus

86236 $\int \frac{d x}{\sqrt{(x - 1) (x - 2)}} =$

1

\log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

2

\log | (x - \frac{3}{2}) - \sqrt{x^{2} - 3 x + 2} | + c

3

\log | (x + \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c

4

\log | (x - 1) + \sqrt{x^{2} - 3 x + 2} | + c

Explanation:

(A) : $I = \int \frac{d x}{\sqrt{(x - 1) (x - 2)}}$
$= \int \frac{d x}{\sqrt{x^{2} - 3 x + 2}}$
$= \int \frac{d x}{\sqrt{x^{2} - 2 \times \frac{3}{2} x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2}}$
$= \int \frac{dx}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}}$
$= \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}} | + c$
$[∵ \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]$
$= \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + c$

MHT CET-2020

Integral Calculus

86237 $\int e^{x} \cdot \sec x (1 + \tan x) d x =$

1

e^{x} \sec x + c

2

e^{x} \tan x + c

3

e^{x} \cos ec x + c

4

e^{x} \cot x + c

Explanation:

(A) : $I = \int e^{x} \cdot \sec x (1 + \tan x) d x$
$= \int e^{x} (\sec x + \sec x \cdot \tan x) d x$
Let, $e^{x} . \sec x = t$
$(e^{x} \cdot \sec x + e^{x} \cdot \sec x \cdot \tan x) d x = d t$
$e^{x} (\sec x + \sec x \cdot \tan x) dx = dt$
$∴ I = \int dt = t + c$
$I = e^{x} \cdot \sec x + c$

MHT CET-2020