Mathematics

# Solve $\displaystyle\int \dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx$

##### SOLUTION
$\begin{array}{l} = \displaystyle\int \dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx\\ = \displaystyle\int \dfrac{{x + 1 - 1}}{{{{\left( {x + 1} \right)}^2}}}dx\\ = \displaystyle\int \left( {\dfrac{1}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx\\ = \displaystyle\int \dfrac{1}{{\left( {x + 1} \right)}}dx - \displaystyle\int \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}dx\\ = \ell n\left( {x + 1} \right) - \dfrac{{{{\left( {x + 1} \right)}^{ - 1}}}}{{ - 1}} + C\\ = \ell n\left( {x + 1} \right) + \dfrac{1}{{\left( {x + 1} \right)}} + C\end{array}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Single Correct Medium
If Y is the solution of $\left ( 1 \, + \, t \right ) \, \,\dfrac{dy}{dt} \, - ty = 1$ and  Y(0)  = -1  then y(1)  is equal to
• A. $e + \frac{-1}{2}$
• B. $e - \frac{-1}{2}$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle\lim_{n \rightarrow \infty} {\dfrac{1}{(n+1)} + \dfrac{1}{(n+2)} + \dfrac{1}{(n+3)} + ... + ... \dfrac{1}{(n+n)}}=?$
• A. $log\ 4$
• B. $log\ 3$
• C. $log\ 5$
• D. $log\ 2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I = \int \frac {dx}{(1 + x^4)^{1/4}}$, then I equals
• A. $\displaystyle \frac {1}{2} \tan^{-1} \left ( \frac {x}{(1 + x^4)^{1/4}} \right ) + C$
• B. $\displaystyle \frac {1}{4} \log \left | \frac {1 - (1 + x^4)^{1/4}}{1 + (1 + x^4)^{1/4}} \right | + C$
• C. $\displaystyle \frac {1}{2} \tan^{-1} \frac {(1 + x^4)^{1/4}}{x} - \frac {1}{4} \log \left | \frac {1 - (1 + x^4)^{1/4}}{1 + (1 + x^4)^{1/4}} \right | + C$
• D. $\displaystyle \frac {1}{2} \tan^{-1} \frac {(1 + x^4)^{1/4}}{x} - \frac {1}{4} \log \left | \frac {x - (1 + x^4)^{1/4}}{x + (1 + x^4)^{1/4}} \right | + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Hard
If $\displaystyle \int x\log \left ( 1+x^{2} \right )dx=\phi \left ( x \right ).\log \left ( 1+x^{2} \right )+\Psi \left ( x \right )+c$ then
• A. $\displaystyle \Psi \left ( x \right )=\frac{1+x^{2}}{2}$
• B. $\displaystyle \phi \left ( x \right )=-\frac{1+x^{2}}{2}$
• C. $\displaystyle \phi \left ( x \right )=\frac{1+x^{2}}{2}$
• D. $\displaystyle \Psi \left ( x \right )=-\frac{1+x^{2}}{2}$

$\displaystyle\int\dfrac{\cos x-\sin x}{\sqrt{\sin 2x}}dx$.