Mathematics

Simplify: $\displaystyle\int \frac{x^{2}}{x^{3}+1}dx$

$\displaystyle \frac{1}{3}\log \left ( 1+x^{3} \right )$

SOLUTION
Let $\displaystyle I=\int \frac { x^{ 2 } }{ x^{ 3 }+1 } dx$

Put $\displaystyle t=x^{ 3 }+1\Rightarrow dt=\frac { { x }^{ 2 } }{ 3 } dx$

Therefore
$\displaystyle I=\frac { 1 }{ 3 } \int { \frac { dt }{ t } } =\frac { 1 }{ 3 } \log { t } =\frac { 1 }{ 3 } \log \left( 1+x^{ 3 } \right)$
Hence, option 'A' is correct.

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

Realted Questions

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If $\int { \cfrac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right) } } dx=\int { \left\{ 1+\cfrac { f(x) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right) } \right\} } dx$
$x+A\tan ^{ -1 }{ \left( \cfrac { x }{ A' } \right) } +B\log { \left( \cfrac { x-l }{ x+m } \right) } +K\quad$ then which of the following is correct
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1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
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If $I=\int \sec^{2} x \mathrm{cosec}^{4}xdx = A\cot^{3}x+B\tan x + C \cot x+D$ then
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
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