Mathematics

# Solve:$\displaystyle \int \dfrac{1}{(x+1)(x^2+1)}dx$

##### SOLUTION
Given the integral,
$\int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) } } dx$
using partial fraction we get,
$\int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) } } dx\\ =\int { [\dfrac { 1 }{ 2(x+1) } -\dfrac { x-1 }{ 2({ x }^{ 2 }+1) } ] } dx\\ =\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ x+1 } } dx-\dfrac { 1 }{ 2 } \int { \dfrac { x-1 }{ { x }^{ 2 }+1 } } dx$
Now, for $\int { \dfrac { 1 }{ x+1 } } dx$
Let, $u=x+1\\ \Rightarrow \dfrac { du }{ dx } =1\\ \Rightarrow du=dx$
Substituting the values of $u$ and $du$ we get,
$\int { \dfrac { 1 }{ x+1 } } dx\\ =\int { \dfrac { 1 }{ u } } du\\ =\ln { (u) } \\ =\ln { (x+1) }$
for $\int { \dfrac { x-1 }{ { x }^{ 2 }+1 } } dx$
Expanding the integral,
$\int { \dfrac { x-1 }{ { x }^{ 2 }+1 } } dx\\ =\int { \dfrac { x }{ { x }^{ 2 }+1 } } dx-\int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx$
Again, for $\int { \dfrac { x }{ { x }^{ 2 }+1 } } dx$
Let,
$u={ x }^{ 2 }+1\\ \Rightarrow \dfrac { du }{ dx } =2x\\ \Rightarrow dx=\dfrac { 1 }{ 2x } du\\ \therefore \int { \dfrac { x }{ { x }^{ 2 }+1 } } dx\\ =\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ u } } du\\ =\dfrac { \ln { (u) } }{ 2 } \\ =\dfrac { \ln { ({ x }^{ 2 }+1) } }{ 2 }$
And
$\int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx\\ =\arctan { (x) } \\ \therefore \int { \dfrac { x }{ { x }^{ 2 }+1 } } dx-\int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx\\ =\dfrac { \ln { ({ x }^{ 2 }+1) } }{ 2 } -\arctan { (x) }$
So,
$\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ x+1 } } dx-\dfrac { 1 }{ 2 } \int { \dfrac { x-1 }{ { x }^{ 2 }+1 } } dx\\ =\dfrac { \ln { (x+1) } }{ 2 } -\dfrac { \ln { ({ x }^{ 2 }+1) } }{ 4 } +\dfrac { \arctan { (x) } }{ 2 } \\ \therefore \int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) } } dx=\dfrac { \ln { (x+1) } }{ 2 } -\dfrac { \ln { ({ x }^{ 2 }+1) } }{ 4 } +\dfrac { \arctan { (x) } }{ 2 } +C.$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
Evaluate : $\displaystyle \int \frac{e^{x-1}+x^{e-1}}{e^{x}+x^{e}}dx.$
• A. $\displaystyle \frac{1}{e}\log \left ( e^{x}-x^{e} \right ).$
• B. $\displaystyle \log \left ( e^{x}+x^{e} \right ).$
• C. $\displaystyle \frac{1}{e^{2}}\log \left ( e^{x}+x^{e} \right ).$
• D. $\displaystyle \frac{1}{e}\log \left ( e^{x}+x^{e} \right ).$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve: $\displaystyle\int\left(\dfrac{x^2-1}{x^2+1}\right)dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
f $k = e^{2007}$ then value of $\displaystyle I =\int_{1}^{k}\frac{ \pi \cos (\pi \log x )} {x} dx$ is
• A. $-\pi$
• B. $\pi/e$
• C. $2007\pi$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve
$\displaystyle \int \dfrac {1-x^{2}}{x(1-2x)} =$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$