Mathematics

# $\displaystyle\int { \dfrac { { x }^{ n-1 } }{ { x }^{ 2n }+{ a }^{ 2 } } dx }$ is equal to

$\dfrac { n }{ a } \tan ^{ -1 }{ \left( \dfrac { { x }^{ n } }{ a } \right) } +C$

##### SOLUTION
Let $I=\displaystyle\int { \dfrac { { x }^{ n-1 } }{ { x }^{ 2n }+{ a }^{ 2 } } dx }$
Put ${ x }^{ n }=t$
$\Rightarrow n{ x }^{ n-1 }dx=dt$
Therefore, $I=\dfrac { 1 }{ n } \displaystyle\int { \dfrac { dt }{ { t }^{ 2 }+{ a }^{ 2 } } }$
$=\dfrac { 1 }{ n } \cdot \dfrac { 1 }{ a } \tan ^{ -1 }{ \dfrac { t }{ a } } +C$
Put $t={ x }^{ n }$
$\Rightarrow I=\dfrac { 1 }{ na } \tan ^{ -1 }{ \left( \dfrac { { x }^{ n } }{ a } \right) } +C$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Subjective Medium
$\displaystyle \int 2x^3+9x^2-8x+5 dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate :
$\int { \dfrac { { x }^{ 2 }\left( x\sec ^{ 2 }{ x } -\tan { x } \right) }{ \left( x\tan { x } -1 \right) ^{ 2 } } }dx =$ ?
• A. $x\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$
• B. $2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x-1}+C$
• C. none
• D. $2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Find the integral of the function    $\displaystyle \frac {1}{\sin x \cos^3 x}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle\int { \dfrac { dx }{ \sqrt { { x }^{ 4 }+{ x }^{ 6 } } } }$ is equal to
• A. $\dfrac { \sqrt { 1+{ x }^{ 2 } } }{ x } +C$
• B. $-\dfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } +C$
• C. $-\dfrac { \sqrt { { x }^{ 2 }-1 } }{ x } +C$
• D. $-\dfrac { \sqrt { 1+{ x }^{ 2 } } }{ x } +C$

Evaluate: $\displaystyle\int {\frac{1}{{{a^x}{b^x}}}} \,dx$