Mathematics

If $y\left( x-y \right) ^{ 2 }=x$, then $\int { \dfrac { 1 }{ x-3y } dx }$ is equal to

$\dfrac { 1 }{ 3 } \log { \left\{ \left( x-y \right) ^{ 2 }+1 \right\} }$

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

Realted Questions

Q1 Subjective Medium
Show that $\displaystyle\int x^{3}\sqrt{q^{2}x^{8}-p^{2}}dx=\frac{1}{4q}\left [ \frac{t}{2}\sqrt{t^{2}-p^{2}}-\frac{1}{2}p^{2}\log \left ( t+\sqrt{t^{2}-p^{2}} \right ) \right ].$ where $t=qx^{4}$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $f(x) = \int^x_1 \dfrac{tan^{-1}t}{t} dt (x=0)$, then the value of $f(e^2)-f(\dfrac{1}{e^2})$
• A. $2\pi$
• B. $\dfrac{\pi}{2}$
• C. $0$
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle\int{\frac{x^2(1-\ln{x})}{\ln^4{x}-x^4}dx}$ is equal to
• A. $\displaystyle\frac{1}{2}\ln{\left(\frac{x}{\ln{x}}\right)}-\frac{1}{4}\ln{(\ln^2{x}-x^2)}+C$
• B. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)+x}{\ln{x}-x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$
• C. $\displaystyle\frac{1}{4}\left(\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}+\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}\right)+C$
• D. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate :
$\int \dfrac{sec^2x}{tanx}dx$

Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$