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Integral Calculus

86144 $\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$ is equal to

1

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

2

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

3

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

4

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

Explanation:

(A) : Given,
$\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$
Let $x = \cos 2 θ$
$θ = \frac{\cos^{- 1} x}{2}$
So, $I = \int \tan^{- 1} \sqrt{(\frac{1 - \cos 2 θ}{1 + \cos 2 θ})} d x$
$= \int \tan^{- 1} \sqrt{\frac{1 - (1 - 2 \sin^{2} θ)}{1 + (2 \cos^{2} θ - 1)}} dx$
$= \int \tan^{- 1} (\tan θ) d x = \int θ . d x$
$= \frac{1}{2} \int \cos^{- 1} xdx = \frac{1}{2} \int 1 \cdot \cos^{- 1} xdx$
$= \frac{1}{2} [x \cdot \cos^{- 1} x - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) \cdot x d x]$
$= \frac{1}{2} [x \cos^{- 1} x + \int \frac{x \cdot dx}{\sqrt{1 - x^{2}}}]$
$= \frac{1}{2} [x \cos^{- 1} x - \sqrt{1 - x^{2}}] + C$

AP EAMCET-2007

Integral Calculus

86145 $\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$ is equal to

1

x + \frac{1}{2} \log (4 \sin x + 6 \cos x) + c

2

2 x + \log (2 \sin x + 3 \cos x) + c

3

x + 2 \log (2 \sin x + 3 \cos x) + c

4

\frac{1}{2} \log (4 \sin x + 6 \cos x) + c

Explanation:

(A) : Given,
$\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$
$\sin x + 8 \cos x = A \frac{d}{d x} (4 \sin x + 6 \cos x) + B (4 \sin x +$ $6 \cos x)$
$\sin x + 8 \cos x = A (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x) B$
We compare the equation and find the value of $A$ and $B -$
$- 6 A + 4 B = 1$
$4 A + 6 B = 8$
So, $A = \frac{1}{2}$ and $B = 1$
$\sin x + 8 \cos x = \frac{1}{2} (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x)$
$\int \frac{(\sin x + 8 \cos x) d x}{4 \sin x + 6 \cos x} = \frac{1}{2} \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} d x + \int \frac{4 \sin x + 6 \cos x}{4 \sin x + 6 \cos x} d x$ Let,
$\Rightarrow 4 \sin x + 6 \cos x = t$
So,
$\Rightarrow (4 \cos x - 6 \sin x) d x = d t$
$= \frac{1}{2} \int \frac{dt}{t} + \int dx = \frac{1}{2} \log | 4 \sin x + 6 \cos x | + x + C$
Hence, $I = x + \frac{1}{2} \log (4 \sin + 6 \cos x) + c$

AP EAMCET-2007

Integral Calculus

86146 If $g (\frac{t + 1}{2 t + 1}) = t + 1$ , then $\int g (x) d x =$

1

\frac{x^{2}}{2} + c

2

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

3

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

4

\frac{x}{2} + \frac{1}{4} \log_{e} | 2 x - 1 | + c

Explanation:

(D) : Given,
Let,
$g (\frac{t + 1}{2 t + 1}) = t + 1$
$\frac{t + 1}{2 t + 1} = x \Rightarrow t + 1 = x (2 t + 1)$
$t - 2 t x = x - 1 \Rightarrow t (1 - 2 x) = x - 1$
$t = \frac{x - 1}{1 - 2 x}$
$So, f (x) = \frac{x - 1}{1 - 2 x} + 1 = \frac{x - 1 + 1 - 2 x}{1 - 2 x} = \frac{x}{2 x - 1}$
$\int \frac{x}{2 x - 1} d x = \frac{1}{2} \int \frac{2 x - 1 + 1}{2 x - 1} d x$
$= \frac{1}{2} [\int \frac{2 x - 1}{2 x - 1} d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [\int d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [x + \frac{1}{2} \log | 2 x - 1 |] + C$
$= \frac{x}{2} + \frac{1}{4} \log | 2 x - 1 | + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86147 $\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x =$

1

\frac{x^{5}}{4} + x^{3} + 6 x^{2} + c

2

\frac{x^{5}}{5} + \frac{x^{4}}{4} + 6 x + c

3

\frac{x^{5}}{5} + x^{3} + 6 x + c

4

\frac{x^{5}}{5} - \frac{x^{3}}{2} + 6 x^{2} + c

Explanation:

(C) : Given,
$\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x$
$x^{8} - 9 x^{2} + 18$ is divided by $x^{4} - 3 x^{2} + 3$ , we get quotient $x^{4} + 3 x^{2} + 6$ and 0 remainder.
Now, $\int (x^{4} + 3 x^{2} + 6) d x$
$= \frac{x^{5}}{5} + \frac{3 x^{3}}{3} + 6 x + C = \frac{x^{5}}{5} + x^{3} + 6 x + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86144 $\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$ is equal to

1

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

2

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

3

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

4

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

Explanation:

(A) : Given,
$\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$
Let $x = \cos 2 θ$
$θ = \frac{\cos^{- 1} x}{2}$
So, $I = \int \tan^{- 1} \sqrt{(\frac{1 - \cos 2 θ}{1 + \cos 2 θ})} d x$
$= \int \tan^{- 1} \sqrt{\frac{1 - (1 - 2 \sin^{2} θ)}{1 + (2 \cos^{2} θ - 1)}} dx$
$= \int \tan^{- 1} (\tan θ) d x = \int θ . d x$
$= \frac{1}{2} \int \cos^{- 1} xdx = \frac{1}{2} \int 1 \cdot \cos^{- 1} xdx$
$= \frac{1}{2} [x \cdot \cos^{- 1} x - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) \cdot x d x]$
$= \frac{1}{2} [x \cos^{- 1} x + \int \frac{x \cdot dx}{\sqrt{1 - x^{2}}}]$
$= \frac{1}{2} [x \cos^{- 1} x - \sqrt{1 - x^{2}}] + C$

AP EAMCET-2007

Integral Calculus

86145 $\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$ is equal to

1

x + \frac{1}{2} \log (4 \sin x + 6 \cos x) + c

2

2 x + \log (2 \sin x + 3 \cos x) + c

3

x + 2 \log (2 \sin x + 3 \cos x) + c

4

\frac{1}{2} \log (4 \sin x + 6 \cos x) + c

Explanation:

(A) : Given,
$\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$
$\sin x + 8 \cos x = A \frac{d}{d x} (4 \sin x + 6 \cos x) + B (4 \sin x +$ $6 \cos x)$
$\sin x + 8 \cos x = A (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x) B$
We compare the equation and find the value of $A$ and $B -$
$- 6 A + 4 B = 1$
$4 A + 6 B = 8$
So, $A = \frac{1}{2}$ and $B = 1$
$\sin x + 8 \cos x = \frac{1}{2} (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x)$
$\int \frac{(\sin x + 8 \cos x) d x}{4 \sin x + 6 \cos x} = \frac{1}{2} \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} d x + \int \frac{4 \sin x + 6 \cos x}{4 \sin x + 6 \cos x} d x$ Let,
$\Rightarrow 4 \sin x + 6 \cos x = t$
So,
$\Rightarrow (4 \cos x - 6 \sin x) d x = d t$
$= \frac{1}{2} \int \frac{dt}{t} + \int dx = \frac{1}{2} \log | 4 \sin x + 6 \cos x | + x + C$
Hence, $I = x + \frac{1}{2} \log (4 \sin + 6 \cos x) + c$

AP EAMCET-2007

Integral Calculus

86146 If $g (\frac{t + 1}{2 t + 1}) = t + 1$ , then $\int g (x) d x =$

1

\frac{x^{2}}{2} + c

2

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

3

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

4

\frac{x}{2} + \frac{1}{4} \log_{e} | 2 x - 1 | + c

Explanation:

(D) : Given,
Let,
$g (\frac{t + 1}{2 t + 1}) = t + 1$
$\frac{t + 1}{2 t + 1} = x \Rightarrow t + 1 = x (2 t + 1)$
$t - 2 t x = x - 1 \Rightarrow t (1 - 2 x) = x - 1$
$t = \frac{x - 1}{1 - 2 x}$
$So, f (x) = \frac{x - 1}{1 - 2 x} + 1 = \frac{x - 1 + 1 - 2 x}{1 - 2 x} = \frac{x}{2 x - 1}$
$\int \frac{x}{2 x - 1} d x = \frac{1}{2} \int \frac{2 x - 1 + 1}{2 x - 1} d x$
$= \frac{1}{2} [\int \frac{2 x - 1}{2 x - 1} d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [\int d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [x + \frac{1}{2} \log | 2 x - 1 |] + C$
$= \frac{x}{2} + \frac{1}{4} \log | 2 x - 1 | + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86147 $\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x =$

1

\frac{x^{5}}{4} + x^{3} + 6 x^{2} + c

2

\frac{x^{5}}{5} + \frac{x^{4}}{4} + 6 x + c

3

\frac{x^{5}}{5} + x^{3} + 6 x + c

4

\frac{x^{5}}{5} - \frac{x^{3}}{2} + 6 x^{2} + c

Explanation:

(C) : Given,
$\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x$
$x^{8} - 9 x^{2} + 18$ is divided by $x^{4} - 3 x^{2} + 3$ , we get quotient $x^{4} + 3 x^{2} + 6$ and 0 remainder.
Now, $\int (x^{4} + 3 x^{2} + 6) d x$
$= \frac{x^{5}}{5} + \frac{3 x^{3}}{3} + 6 x + C = \frac{x^{5}}{5} + x^{3} + 6 x + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86144 $\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$ is equal to

1

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

2

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

3

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

4

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

Explanation:

(A) : Given,
$\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$
Let $x = \cos 2 θ$
$θ = \frac{\cos^{- 1} x}{2}$
So, $I = \int \tan^{- 1} \sqrt{(\frac{1 - \cos 2 θ}{1 + \cos 2 θ})} d x$
$= \int \tan^{- 1} \sqrt{\frac{1 - (1 - 2 \sin^{2} θ)}{1 + (2 \cos^{2} θ - 1)}} dx$
$= \int \tan^{- 1} (\tan θ) d x = \int θ . d x$
$= \frac{1}{2} \int \cos^{- 1} xdx = \frac{1}{2} \int 1 \cdot \cos^{- 1} xdx$
$= \frac{1}{2} [x \cdot \cos^{- 1} x - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) \cdot x d x]$
$= \frac{1}{2} [x \cos^{- 1} x + \int \frac{x \cdot dx}{\sqrt{1 - x^{2}}}]$
$= \frac{1}{2} [x \cos^{- 1} x - \sqrt{1 - x^{2}}] + C$

AP EAMCET-2007

Integral Calculus

86145 $\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$ is equal to

1

x + \frac{1}{2} \log (4 \sin x + 6 \cos x) + c

2

2 x + \log (2 \sin x + 3 \cos x) + c

3

x + 2 \log (2 \sin x + 3 \cos x) + c

4

\frac{1}{2} \log (4 \sin x + 6 \cos x) + c

Explanation:

(A) : Given,
$\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$
$\sin x + 8 \cos x = A \frac{d}{d x} (4 \sin x + 6 \cos x) + B (4 \sin x +$ $6 \cos x)$
$\sin x + 8 \cos x = A (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x) B$
We compare the equation and find the value of $A$ and $B -$
$- 6 A + 4 B = 1$
$4 A + 6 B = 8$
So, $A = \frac{1}{2}$ and $B = 1$
$\sin x + 8 \cos x = \frac{1}{2} (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x)$
$\int \frac{(\sin x + 8 \cos x) d x}{4 \sin x + 6 \cos x} = \frac{1}{2} \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} d x + \int \frac{4 \sin x + 6 \cos x}{4 \sin x + 6 \cos x} d x$ Let,
$\Rightarrow 4 \sin x + 6 \cos x = t$
So,
$\Rightarrow (4 \cos x - 6 \sin x) d x = d t$
$= \frac{1}{2} \int \frac{dt}{t} + \int dx = \frac{1}{2} \log | 4 \sin x + 6 \cos x | + x + C$
Hence, $I = x + \frac{1}{2} \log (4 \sin + 6 \cos x) + c$

AP EAMCET-2007

Integral Calculus

86146 If $g (\frac{t + 1}{2 t + 1}) = t + 1$ , then $\int g (x) d x =$

1

\frac{x^{2}}{2} + c

2

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

3

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

4

\frac{x}{2} + \frac{1}{4} \log_{e} | 2 x - 1 | + c

Explanation:

(D) : Given,
Let,
$g (\frac{t + 1}{2 t + 1}) = t + 1$
$\frac{t + 1}{2 t + 1} = x \Rightarrow t + 1 = x (2 t + 1)$
$t - 2 t x = x - 1 \Rightarrow t (1 - 2 x) = x - 1$
$t = \frac{x - 1}{1 - 2 x}$
$So, f (x) = \frac{x - 1}{1 - 2 x} + 1 = \frac{x - 1 + 1 - 2 x}{1 - 2 x} = \frac{x}{2 x - 1}$
$\int \frac{x}{2 x - 1} d x = \frac{1}{2} \int \frac{2 x - 1 + 1}{2 x - 1} d x$
$= \frac{1}{2} [\int \frac{2 x - 1}{2 x - 1} d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [\int d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [x + \frac{1}{2} \log | 2 x - 1 |] + C$
$= \frac{x}{2} + \frac{1}{4} \log | 2 x - 1 | + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86147 $\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x =$

1

\frac{x^{5}}{4} + x^{3} + 6 x^{2} + c

2

\frac{x^{5}}{5} + \frac{x^{4}}{4} + 6 x + c

3

\frac{x^{5}}{5} + x^{3} + 6 x + c

4

\frac{x^{5}}{5} - \frac{x^{3}}{2} + 6 x^{2} + c

Explanation:

(C) : Given,
$\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x$
$x^{8} - 9 x^{2} + 18$ is divided by $x^{4} - 3 x^{2} + 3$ , we get quotient $x^{4} + 3 x^{2} + 6$ and 0 remainder.
Now, $\int (x^{4} + 3 x^{2} + 6) d x$
$= \frac{x^{5}}{5} + \frac{3 x^{3}}{3} + 6 x + C = \frac{x^{5}}{5} + x^{3} + 6 x + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86144 $\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$ is equal to

1

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

2

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

3

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

4

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

Explanation:

(A) : Given,
$\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) d x$
Let $x = \cos 2 θ$
$θ = \frac{\cos^{- 1} x}{2}$
So, $I = \int \tan^{- 1} \sqrt{(\frac{1 - \cos 2 θ}{1 + \cos 2 θ})} d x$
$= \int \tan^{- 1} \sqrt{\frac{1 - (1 - 2 \sin^{2} θ)}{1 + (2 \cos^{2} θ - 1)}} dx$
$= \int \tan^{- 1} (\tan θ) d x = \int θ . d x$
$= \frac{1}{2} \int \cos^{- 1} xdx = \frac{1}{2} \int 1 \cdot \cos^{- 1} xdx$
$= \frac{1}{2} [x \cdot \cos^{- 1} x - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) \cdot x d x]$
$= \frac{1}{2} [x \cos^{- 1} x + \int \frac{x \cdot dx}{\sqrt{1 - x^{2}}}]$
$= \frac{1}{2} [x \cos^{- 1} x - \sqrt{1 - x^{2}}] + C$

AP EAMCET-2007

Integral Calculus

86145 $\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$ is equal to

1

x + \frac{1}{2} \log (4 \sin x + 6 \cos x) + c

2

2 x + \log (2 \sin x + 3 \cos x) + c

3

x + 2 \log (2 \sin x + 3 \cos x) + c

4

\frac{1}{2} \log (4 \sin x + 6 \cos x) + c

Explanation:

(A) : Given,
$\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} d x$
$\sin x + 8 \cos x = A \frac{d}{d x} (4 \sin x + 6 \cos x) + B (4 \sin x +$ $6 \cos x)$
$\sin x + 8 \cos x = A (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x) B$
We compare the equation and find the value of $A$ and $B -$
$- 6 A + 4 B = 1$
$4 A + 6 B = 8$
So, $A = \frac{1}{2}$ and $B = 1$
$\sin x + 8 \cos x = \frac{1}{2} (4 \cos x - 6 \sin x) + (4 \sin x + 6 \cos x)$
$\int \frac{(\sin x + 8 \cos x) d x}{4 \sin x + 6 \cos x} = \frac{1}{2} \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} d x + \int \frac{4 \sin x + 6 \cos x}{4 \sin x + 6 \cos x} d x$ Let,
$\Rightarrow 4 \sin x + 6 \cos x = t$
So,
$\Rightarrow (4 \cos x - 6 \sin x) d x = d t$
$= \frac{1}{2} \int \frac{dt}{t} + \int dx = \frac{1}{2} \log | 4 \sin x + 6 \cos x | + x + C$
Hence, $I = x + \frac{1}{2} \log (4 \sin + 6 \cos x) + c$

AP EAMCET-2007

Integral Calculus

86146 If $g (\frac{t + 1}{2 t + 1}) = t + 1$ , then $\int g (x) d x =$

1

\frac{x^{2}}{2} + c

2

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

3

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

4

\frac{x}{2} + \frac{1}{4} \log_{e} | 2 x - 1 | + c

Explanation:

(D) : Given,
Let,
$g (\frac{t + 1}{2 t + 1}) = t + 1$
$\frac{t + 1}{2 t + 1} = x \Rightarrow t + 1 = x (2 t + 1)$
$t - 2 t x = x - 1 \Rightarrow t (1 - 2 x) = x - 1$
$t = \frac{x - 1}{1 - 2 x}$
$So, f (x) = \frac{x - 1}{1 - 2 x} + 1 = \frac{x - 1 + 1 - 2 x}{1 - 2 x} = \frac{x}{2 x - 1}$
$\int \frac{x}{2 x - 1} d x = \frac{1}{2} \int \frac{2 x - 1 + 1}{2 x - 1} d x$
$= \frac{1}{2} [\int \frac{2 x - 1}{2 x - 1} d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [\int d x + \int \frac{1}{2 x - 1} d x]$
$= \frac{1}{2} [x + \frac{1}{2} \log | 2 x - 1 |] + C$
$= \frac{x}{2} + \frac{1}{4} \log | 2 x - 1 | + C$

AP EAMCET-2018-24.04.2018

Integral Calculus

86147 $\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x =$

1

\frac{x^{5}}{4} + x^{3} + 6 x^{2} + c

2

\frac{x^{5}}{5} + \frac{x^{4}}{4} + 6 x + c

3

\frac{x^{5}}{5} + x^{3} + 6 x + c

4

\frac{x^{5}}{5} - \frac{x^{3}}{2} + 6 x^{2} + c

Explanation:

(C) : Given,
$\int \frac{x^{8} - 9 x^{2} + 18}{x^{4} - 3 x^{2} + 3} d x$
$x^{8} - 9 x^{2} + 18$ is divided by $x^{4} - 3 x^{2} + 3$ , we get quotient $x^{4} + 3 x^{2} + 6$ and 0 remainder.
Now, $\int (x^{4} + 3 x^{2} + 6) d x$
$= \frac{x^{5}}{5} + \frac{3 x^{3}}{3} + 6 x + C = \frac{x^{5}}{5} + x^{3} + 6 x + C$

AP EAMCET-2018-24.04.2018

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