∫ dx/(x+1)²(x²+1) =

Correct Answer is : log⁡ex+1_e{x+1}logex+1 - 12log⁡ex2+11/2_e{x²+1}21logex2+1 - 12(x+1)1/2(x+1)2(x+1)1 + c

_e{x+1} + 1/2_e{x²+1} - 1/x+1 + c

_e{x+1} - 1/2_e{x²+1} - 1/2(x+1) + c

1/2_e{x+1} - 1/4_e{x²+1} + 1/2(x+1) + c

1/4_e{x+1} + 1/2_e{x²+1} - 1/x+1 + c

Correct Answer is : log⁡ex+1_e{x+1}logex+1 - 12log⁡ex2+11/2_e{x²+1}21logex2+1 - 12(x+1)1/2(x+1)2(x+1)1 + c

Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

\frac{dx}{(x+1)^2(x^2+1)}

$\log_e\sqrt{x+1}$ + $\frac{1}{2}\log_e\sqrt{x^2+1}$ - $\frac{1}{x+1}$ + c
$\log_e\sqrt{x+1}$ - $\frac{1}{2}\log_e\sqrt{x^2+1}$ - $\frac{1}{2(x+1)}$ + c
$\frac{1}{2}\log_e\sqrt{x+1}$ - $\frac{1}{4}\log_e\sqrt{x^2+1}$ + $\frac{1}{2(x+1)}$ + c
$\frac{1}{4}\log_e\sqrt{x+1}$ + $\frac{1}{2}\log_e\sqrt{x^2+1}$ - $\frac{1}{x+1}$ + c

Solution:

The integral expression is given as,

I=\int\;\frac{dx}{(x+1)^2(x^2+1)}

Consider

\;\frac{1}{(x+1)^2(x^2+1)} \;=\; \frac{A}{(x+1)^2} + \;\frac{B}{(x^2+1)} + \;\frac{Cx+D}{x^2+1}

1=A(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x+1)^2

Substitute

x=-1

A=\;\frac{1}{2}

Substitute

x^2=-1

1=(Cx+D)(x^2+2x+1)

1=(Cx+D)2x

1=-2C+2Dx

Substitute

x=0

-\;\frac{1}{2}=C, D=0

Comparing coefficient of

x^3

0=B+C

B=\;\frac{1}{2}

Now

I=\int\left[\;\frac{1}{2}\;\frac{1}{(x+1)^2}+\;\frac{1}{2}\;\frac{1}{x+1}-\;\frac{1}{2}\;\frac{x}{x^2+1}\right]dx

=\;\frac{1}{2}\left[-\;\frac{1}{x+1}+\log(x+1)-\;\frac{1}{2}\log(x^2+1)\right]+c

=\log\sqrt{x+1}-\;\frac{1}{2}\log\sqrt{x^2+1}-\;\frac{1}{2(x+1)}+c

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