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Integral Calculus

86243 $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x =$

1

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

2

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

3

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

4

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

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 1} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) d x = d t$
$∴ I = \int \frac{dt}{t^{2} + 1} = \tan^{- 1} (t) + c$
$= \tan^{- 1} (x - \frac{1}{x}) + c = \tan^{- 1} (\frac{x^{2} - 1}{x}) + c$

MHT CET-2019

Integral Calculus

86244 $\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x =$

1

a - b

2

\frac{b - a}{2}

3

\frac{a - b}{2}

4

a + b

Explanation:

(B) : $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x}} d x I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} d x$
From equation (i) and equation (ii) adding, we get -
$2 I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$2 I = \int_{a}^{b} d x = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

MHT CET-2019

Integral Calculus

86245 If $\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C$ then $A + B =$

1

\frac{9}{4}

2

\frac{19}{4}

3

\frac{3}{4}

4

\frac{13}{12}

Explanation:

(D) : Given, $\int \frac{dx}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (Bx) + C$
L.H.S., $I = \int \frac{dx}{\sqrt{16 - 9 x^{2}}}$
$= \frac{1}{3} \int \frac{dx}{\sqrt{{(\frac{4}{3})}^{2} - x^{2}}} = \frac{1}{3} \sin^{- 1} \frac{3 x}{4} + C$
On comparing L.H.S. and R.H.S., we get-
$A = \frac{1}{3}, B = \frac{3}{4}$
$∴ A + B = \frac{1}{3} + \frac{3}{4} = \frac{13}{12}$

MHT CET-2018

Integral Calculus

86246 $\int e^{x} [\frac{2 + \sin 2 x}{1 + \cos 2 x}] d x =$

1

e^{x} \tan x + c

2

e^{x} + \tan x + c

3

2 e^{x} \tan x + c

4

e^{x} \tan 2 x + c

Explanation:

(A) : $I = \int e^{x} (\frac{2 + \sin 2 x}{1 + \cos 2 x}) d x$
$= \int e^{x} (\frac{2}{1 + \cos 2 x} + \frac{\sin 2 x}{1 + \cos 2 x}) dx$ $= \int e^{x} (\frac{2}{2 \cos^{2} x} + \frac{2 \sin x \cdot \cos x}{2 \cos^{2} x}) dx$
$= \int e^{x} (\sec^{2} x + \tan x) dx$
$[∵ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c]$
$= e^{x} \cdot \tan x + c$

MHT CET-2018

Integral Calculus

86243 $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x =$

1

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

2

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

3

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

4

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

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 1} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) d x = d t$
$∴ I = \int \frac{dt}{t^{2} + 1} = \tan^{- 1} (t) + c$
$= \tan^{- 1} (x - \frac{1}{x}) + c = \tan^{- 1} (\frac{x^{2} - 1}{x}) + c$

MHT CET-2019

Integral Calculus

86244 $\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x =$

1

a - b

2

\frac{b - a}{2}

3

\frac{a - b}{2}

4

a + b

Explanation:

(B) : $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x}} d x I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} d x$
From equation (i) and equation (ii) adding, we get -
$2 I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$2 I = \int_{a}^{b} d x = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

MHT CET-2019

Integral Calculus

86245 If $\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C$ then $A + B =$

1

\frac{9}{4}

2

\frac{19}{4}

3

\frac{3}{4}

4

\frac{13}{12}

Explanation:

(D) : Given, $\int \frac{dx}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (Bx) + C$
L.H.S., $I = \int \frac{dx}{\sqrt{16 - 9 x^{2}}}$
$= \frac{1}{3} \int \frac{dx}{\sqrt{{(\frac{4}{3})}^{2} - x^{2}}} = \frac{1}{3} \sin^{- 1} \frac{3 x}{4} + C$
On comparing L.H.S. and R.H.S., we get-
$A = \frac{1}{3}, B = \frac{3}{4}$
$∴ A + B = \frac{1}{3} + \frac{3}{4} = \frac{13}{12}$

MHT CET-2018

Integral Calculus

86246 $\int e^{x} [\frac{2 + \sin 2 x}{1 + \cos 2 x}] d x =$

1

e^{x} \tan x + c

2

e^{x} + \tan x + c

3

2 e^{x} \tan x + c

4

e^{x} \tan 2 x + c

Explanation:

(A) : $I = \int e^{x} (\frac{2 + \sin 2 x}{1 + \cos 2 x}) d x$
$= \int e^{x} (\frac{2}{1 + \cos 2 x} + \frac{\sin 2 x}{1 + \cos 2 x}) dx$ $= \int e^{x} (\frac{2}{2 \cos^{2} x} + \frac{2 \sin x \cdot \cos x}{2 \cos^{2} x}) dx$
$= \int e^{x} (\sec^{2} x + \tan x) dx$
$[∵ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c]$
$= e^{x} \cdot \tan x + c$

MHT CET-2018

Integral Calculus

86243 $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x =$

1

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

2

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

3

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

4

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

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 1} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) d x = d t$
$∴ I = \int \frac{dt}{t^{2} + 1} = \tan^{- 1} (t) + c$
$= \tan^{- 1} (x - \frac{1}{x}) + c = \tan^{- 1} (\frac{x^{2} - 1}{x}) + c$

MHT CET-2019

Integral Calculus

86244 $\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x =$

1

a - b

2

\frac{b - a}{2}

3

\frac{a - b}{2}

4

a + b

Explanation:

(B) : $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x}} d x I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} d x$
From equation (i) and equation (ii) adding, we get -
$2 I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$2 I = \int_{a}^{b} d x = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

MHT CET-2019

Integral Calculus

86245 If $\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C$ then $A + B =$

1

\frac{9}{4}

2

\frac{19}{4}

3

\frac{3}{4}

4

\frac{13}{12}

Explanation:

(D) : Given, $\int \frac{dx}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (Bx) + C$
L.H.S., $I = \int \frac{dx}{\sqrt{16 - 9 x^{2}}}$
$= \frac{1}{3} \int \frac{dx}{\sqrt{{(\frac{4}{3})}^{2} - x^{2}}} = \frac{1}{3} \sin^{- 1} \frac{3 x}{4} + C$
On comparing L.H.S. and R.H.S., we get-
$A = \frac{1}{3}, B = \frac{3}{4}$
$∴ A + B = \frac{1}{3} + \frac{3}{4} = \frac{13}{12}$

MHT CET-2018

Integral Calculus

86246 $\int e^{x} [\frac{2 + \sin 2 x}{1 + \cos 2 x}] d x =$

1

e^{x} \tan x + c

2

e^{x} + \tan x + c

3

2 e^{x} \tan x + c

4

e^{x} \tan 2 x + c

Explanation:

(A) : $I = \int e^{x} (\frac{2 + \sin 2 x}{1 + \cos 2 x}) d x$
$= \int e^{x} (\frac{2}{1 + \cos 2 x} + \frac{\sin 2 x}{1 + \cos 2 x}) dx$ $= \int e^{x} (\frac{2}{2 \cos^{2} x} + \frac{2 \sin x \cdot \cos x}{2 \cos^{2} x}) dx$
$= \int e^{x} (\sec^{2} x + \tan x) dx$
$[∵ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c]$
$= e^{x} \cdot \tan x + c$

MHT CET-2018

Integral Calculus

86243 $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x =$

1

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

2

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

3

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

4

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

Explanation:

(A) : $I = \int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$
$= \int \frac{1 + \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 1} d x$
Let, $x - \frac{1}{x} = t$
$(1 + \frac{1}{x^{2}}) d x = d t$
$∴ I = \int \frac{dt}{t^{2} + 1} = \tan^{- 1} (t) + c$
$= \tan^{- 1} (x - \frac{1}{x}) + c = \tan^{- 1} (\frac{x^{2} - 1}{x}) + c$

MHT CET-2019

Integral Calculus

86244 $\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x =$

1

a - b

2

\frac{b - a}{2}

3

\frac{a - b}{2}

4

a + b

Explanation:

(B) : $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x}} d x I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} d x$
From equation (i) and equation (ii) adding, we get -
$2 I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} d x$
$2 I = \int_{a}^{b} d x = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

MHT CET-2019

Integral Calculus

86245 If $\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C$ then $A + B =$

1

\frac{9}{4}

2

\frac{19}{4}

3

\frac{3}{4}

4

\frac{13}{12}

Explanation:

(D) : Given, $\int \frac{dx}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (Bx) + C$
L.H.S., $I = \int \frac{dx}{\sqrt{16 - 9 x^{2}}}$
$= \frac{1}{3} \int \frac{dx}{\sqrt{{(\frac{4}{3})}^{2} - x^{2}}} = \frac{1}{3} \sin^{- 1} \frac{3 x}{4} + C$
On comparing L.H.S. and R.H.S., we get-
$A = \frac{1}{3}, B = \frac{3}{4}$
$∴ A + B = \frac{1}{3} + \frac{3}{4} = \frac{13}{12}$

MHT CET-2018

Integral Calculus

86246 $\int e^{x} [\frac{2 + \sin 2 x}{1 + \cos 2 x}] d x =$

1

e^{x} \tan x + c

2

e^{x} + \tan x + c

3

2 e^{x} \tan x + c

4

e^{x} \tan 2 x + c

Explanation:

(A) : $I = \int e^{x} (\frac{2 + \sin 2 x}{1 + \cos 2 x}) d x$
$= \int e^{x} (\frac{2}{1 + \cos 2 x} + \frac{\sin 2 x}{1 + \cos 2 x}) dx$ $= \int e^{x} (\frac{2}{2 \cos^{2} x} + \frac{2 \sin x \cdot \cos x}{2 \cos^{2} x}) dx$
$= \int e^{x} (\sec^{2} x + \tan x) dx$
$[∵ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c]$
$= e^{x} \cdot \tan x + c$

MHT CET-2018

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