Mathematics

# $I= \int \frac{x+2}{(x+1)^2}dx;$ then I is equal to

$\log (1+x)-\dfrac{1}{x+1}+c$

##### SOLUTION
$\int { \dfrac { x+2 }{ { \left( x+1 \right) }^{ 2 } } } dx$
$=\int { \dfrac { x+1+1 }{ { \left( x+1 \right) }^{ 2 } } } dx$
$=\int { \left( \dfrac { x+1 }{ { \left( x+1 \right) }^{ 2 } } +\dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } \right) } dx$
$=\int { \dfrac { 1 }{ x+1 } } dx+\int { \dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } } dx$            $\left[ \because \int { { x }^{ n }dx=\dfrac { { x }^{ n+1 } }{ n+1 } +c } \right]$

$=\log \left(x+1\right) +\dfrac { { \left( x+1 \right) }^{ -2+1 } }{ -2+1 } +c$       $\left[ \because \int { \dfrac { 1 }{ x } dx=\log { x+c } } \right]$
$=\log\left(x+1\right) +\left(-1\right) \dfrac{1}{\left(x+1\right)}+c$
$=\log\left(x+1\right)-\dfrac{1}{x+1}+c$
Hence, the answer is $\log\left(x+1\right)-\dfrac{1}{x+1}+c.$

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

#### Realted Questions

Q1 Single Correct Medium
If $\Delta (x)=\left| \begin{matrix} 1+x+2{ x }^{ 2 } & x+3 & 1 \\ x+2{ x }^{ 2 } & x & 3 \\ 3x+6{ x }^{ 2 } & 3x+11 & 9 \end{matrix} \right|$ then $\displaystyle \int^{1}_{0}\Delta (x)dx$ is
• A. $-\dfrac {176}{3}$
• B. $\dfrac {186}{3}$
• C. $-\dfrac {192}{3}$
• D. $\dfrac {176}{5}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int\frac{\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}dx=$
• A. $\log[\sin^{-1}x+\cos^{-1}x]+c$
• B. $\dfrac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]+c$
• C. $\dfrac{4}{\pi}[x\sin^{-1}x-\sqrt{1-x^{2}}]+C$
• D. $\displaystyle \frac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]-x+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral:
$\displaystyle\int{\sqrt{\dfrac{1-\sin{2x}}{1+\sin{2x}}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate : $\displaystyle\int _ { 1 } ^ { 2 } \frac { \ln x } { x ^ { 2 } } d x$

Consider the integrals $I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tan x}}$ and $I_2=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{sin x}dx}{\sqrt{sin }x+\sqrt{cos}x}$