Mathematics

# $\int {\dfrac{{{x^5}}}{{{x^2}\, + \,9}}} \,dx$

##### SOLUTION

Consider the given integral.

$I=\int{\dfrac{{{x}^{5}}}{{{x}^{2}}+9}}dx$

Let,

$t={{x}^{2}}+9$

$dt=2xdx$

Therefore,

$I=\dfrac{1}{2}\int{\dfrac{{{\left( t-9 \right)}^{2}}}{t}}dt$

$I=\dfrac{1}{2}\int{\dfrac{{{t}^{2}}+81-18t}{t}}dt$

$I=\dfrac{1}{2}\int{\left( t+\dfrac{81}{t}-18 \right)}dt$

$I=\dfrac{1}{2}\left[ \dfrac{{{t}^{2}}}{2}+81\ln \left( t \right)-18t \right]+C$

Put the value of $t$, we get

$I=\dfrac{1}{2}\left[ \dfrac{{{\left( {{x}^{2}}+9 \right)}^{2}}}{2}+81\ln \left( {{x}^{2}}+9 \right)-18\left( {{x}^{2}}+9 \right) \right]+C$

Hence, the value of this expression is $\dfrac{1}{2}\left[ \dfrac{{{\left( {{x}^{2}}+9 \right)}^{2}}}{2}+81\ln \left( {{x}^{2}}+9 \right)-18\left( {{x}^{2}}+9 \right) \right]+C$

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

#### Realted Questions

Q1 Multiple Correct Hard
If $\displaystyle \int \frac{4e^{x}+6e^{-x}}{9e^{x}-4e^{-x}}dx=Ax+B\log _{e}\left ( 9e^{2x}-4 \right )+C$ then
• A. $\displaystyle A=\frac{3}{2}$
• B. $\displaystyle B=\frac{35}{36}$
• C. C is indefinite
• D. $\displaystyle A+B=\frac{-19}{36}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\int x ^ { 2 } \cdot \cos \left( x ^ { 3 } \right) \sqrt { \sin ^ { 7 } \left( x ^ { 3 } \right) } \cdot d x$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int \cos^{-1}\sqrt{\frac{1-x}{2}}dx=$
• A. $\displaystyle \frac{\pi}{2}x-\frac{1}{2}xcos^{-1}x+c$
• B. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x-\sqrt{1-x^{2}}+c$
• C. $\displaystyle \dfrac{\pi}{2}x+\dfrac{1}{2}xcos^{-1}x+c$
• D. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x+\dfrac{1}{2}\sqrt{1-x^{2}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve $\displaystyle\int \dfrac {2\cos x-3\sin x}{6\cos x+4\sin x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
If $\int _{ 0 }^{ a }{ \quad } \frac { dx }{ \sqrt { x+a } +\sqrt { x } } =\int _{ 0 }^{ \pi /8 }{ \frac { 2\tan { \theta } }{ \sin { 2\theta } } d\theta } ,\quad then\quad value\quad of\quad 'a'\quad is\quad equal\quad to\quad \left( a\quad >\quad 0 \right)$
• A. $\frac { \pi }{ 4 }$
• B. $\frac { 3\pi }{ 4 }$
• C. $\frac { 9 }{ 16 }$
• D. $\frac { 3 }{ 4 }$