Mathematics

# $\displaystyle\int \frac{x^{3}}{25x^{8}-16}dx.$

$\displaystyle \frac{1}{160}\log \frac{5x^{4}-4}{5x^{4}+4}$

##### SOLUTION
Let $\displaystyle I=\int \frac { x^{ 3 } }{ 25x^{ 8 }-16 } dx$
Put $\displaystyle 5x^{ 4 }=t\Rightarrow 20x^{ 3 }dx=dt$

$\displaystyle I=\frac { 1 }{ 20 } \int \frac { dt }{ t^{ 2 }-4^{ 2 } } =\frac { 1 }{ 20 } .\frac { 1 }{ 2.4 } \log \frac { t-4 }{ t+4 }$

$\displaystyle =\frac { 1 }{ 160 } \log \frac { 5x^{ 4 }-4 }{ 5x^{ 4 }+4 }$
Hence, option 'C' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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