Mathematics

# Solve:$\int\displaystyle\dfrac{x}{x^4-a^4}$

##### SOLUTION
$\int { \dfrac { x }{ { x }^{ 4 }-{ a }^{ 4 } } dx } \\ Let\quad { x }^{ 2 }=t\\ differentiating\quad w.r.t\quad x,\quad we\quad get\\ 2xdx=dt\quad \\ Now,\\ \int { \dfrac { x }{ { x }^{ 4 }-{ a }^{ 4 } } dx } \\ =\int { \dfrac { \dfrac { 1 }{ 2 } dt }{ { t }^{ 2 }-({ a }^{ 2 })^{ 2 } } } \\ =\dfrac { 1 }{ 2 } \int { \dfrac { dt }{ { t }^{ 2 }-({ a }^{ 2 })^{ 2 } } } \\ =\dfrac { 1 }{ 2 } \left[ \dfrac { 1 }{ { 2a }^{ 2 } } \log\left| \dfrac { t-{ a }^{ 2 } }{ t-{ a }^{ 2 } } \right| \right] +C\\ =\dfrac { 1 }{ { 4a }^{ 2 } } \log\left| \dfrac { x-{ a }^{ 2 } }{ x-{ a }^{ 2 } } \right| +C$

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

#### Realted Questions

Q1 Subjective Hard
Evaluate $\int_{0}^{4}(|x| + |x - 2| + |x - 4|)$dx.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium

$I =\int \frac{ ( x + x^{} + x^{}) }{ x(1+x^{}) } dx$
equals to

• A. (b)$\frac{3}{2} x^{} - 6 \ tan^{-1} ( x^{} ) + c$
• B. (c)$\frac{3}{2} x^{} + \ tan^{-1} ( x^{} ) + c$
• C. None of these
• D. (a)$\frac{3}{2} x^{} + 6 \ tan^{-1} ( x^{} ) + c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium

If $f(x)=\displaystyle \frac{e^{x}}{1+e^{x}}$ ,$I_{1}=\displaystyle \int_{f(-a)}^{f(a)}xg\{x(1-x)\}dx$ and $I_{2}=\displaystyle \int_{f(-a)}^{f(a)}g\{x(1-x)\}dx$, then the value $\displaystyle \frac{I_{2}}{I_{1}}$ is
• A. -3
• B. -1
• C. 1
• D. 2

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate: $\displaystyle \int (x^2+5x+6)(2x+5)d x$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Medium
Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$

On the basis of above information, answer the following questions:

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020