Mathematics

# Solve $\displaystyle \int{\dfrac{2x\ln \left( {{x}^{2}}-1 \right)}{\left( {{x}^{2}}-1 \right)}}dx$

##### SOLUTION

Consider the given integral.

$I=\displaystyle\int{\dfrac{2x\ln \left( {{x}^{2}}-1 \right)}{\left( {{x}^{2}}-1 \right)}}dx$

Let $t=\ln \left( {{x}^{2}}-1 \right)$

$\dfrac{dt}{dx}=\dfrac{1}{{{x}^{2}}-1}\times 2x$

$dt=\dfrac{2x}{{{x}^{2}}-1}dx$

Therefore,

$I=\displaystyle\int{tdt}$

$I=\dfrac{{{t}^{2}}}{2}+C$

On putting the value of $t$, we get

$I=\dfrac{{{\left( \ln \left( {{x}^{2}}-1 \right) \right)}^{2}}}{2}+C$

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

#### Realted Questions

Q1 Subjective Medium

Evaluate the following definite integral:

$\displaystyle\int_{1}^{2} \dfrac 2x\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate the following integrals:
$\int \dfrac{x}{1+x^{2}} dx$, $x \epsilon R$
• A. $\dfrac{1}{3} log(1+x^{3}) +C$
• B. $tan^{-1}x+C$
• C. None of these
• D. $\dfrac{1}{2} log(1+x^{2}) +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Prove that:
$\displaystyle \int \dfrac {x^{2}dx}{(x\sin x+\cos x)^{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, }$ then $I$ equals
• A. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
• B. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 2 }\sqrt { x } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
• C. None of these
• D. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$

solve $\int \frac{1}{1+e^{-1}}\;dx$