Mathematics

# $\displaystyle \int \dfrac {x^2}{x^3+64} dx$

##### SOLUTION
$\displaystyle \int \dfrac{x^2}{x^3+64}dx\\t=x^3+64\implies dt=3x^2dt\\\displaystyle \dfrac 13\int \dfrac 1t dt=\dfrac 13\log t+c$

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

#### Realted Questions

Q1 Single Correct Hard

lf $f(x)=p(x)q(x)$ where $p(x)=\sqrt{\cos x},\ q(x)=\displaystyle \log(\frac{1-x}{1+x})$ then $\displaystyle \int^{b}_{a} f(x)dx$ equals where $a=-1/2 b=1/2$
• A. $2\displaystyle \int_{0}^{1/2}p(x)q(x)dx$
• B. $\displaystyle \int_{0}^{b}p(x)q(x)$
• C. 1
• D.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle I = \int { \frac { \cos { x } }{ 1-\sin { x } \cos { x } } dx },$ then $I$ equals
• A. $\displaystyle \tan ^{ -1 }{ \left( \sin { x } -\cos { x } \right) } +\frac { 1 }{ 2\sqrt { 3 } } \ln { \left| \frac { \sin { x } +\cos { x } -\sqrt { 3 } }{ \sin { x } +\cos { x } +\sqrt { 3 } } \right| } +c$
• B. $\displaystyle \tan ^{ -1 }{ \left( \sin { x } -\cos { x } \right) } -\frac { 1 }{ \sqrt { 3 } } \ln { \left| \frac { \sin { x } +\cos { x } -\sqrt { 3 } }{ \sin { x } +\cos { x } +\sqrt { 3 } } \right| } +c$
• C. $\displaystyle \tan ^{ -1 }{ \left( \sin { x } -\cos { x } \right) } +\frac { 1 }{ \sqrt { 3 } } \ln { \left| \frac { \sin { x } +\cos { x } -\sqrt { 3 } }{ \sin { x } +\cos { x } +\sqrt { 3 } } \right| } +c$
• D. $\displaystyle \tan ^{ -1 }{ \left( \sin { x } -\cos { x } \right) } -\frac { 1 }{ 2\sqrt { 3 } } \ln { \left| \frac { \sin { x } +\cos { x } -\sqrt { 3 } }{ \sin { x } +\cos { x } +\sqrt { 3 } } \right| } +c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve:-
$\int {\frac{{dx}}{{\sqrt {{e^{2x}} - 1} }}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$n\overset{Lt}{\rightarrow}\infty \displaystyle \frac{[1+4+9+\ldots+n^{2}]}{n^{3}}=$
• A. 2
• B. 3
• C. 1
• D. 1/3

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$