Mathematics

# Evaluate : $\displaystyle \int \dfrac{(x^{2}+1)e^{x}}{(x+1)^{2}} dx$

##### SOLUTION
Let $I=\displaystyle\int{{e}^{x}\left(\dfrac{{x}^{2}+1}{{\left(x+1\right)}^{2}}\right)dx}$
$=\displaystyle\int{{e}^{x}\left(\dfrac{{x}^{2}-1+2}{{\left(x+1\right)}^{2}}\right)dx}$
$=\displaystyle\int{{e}^{x}\left(\dfrac{\left(x-1\right)\left(x+1\right)+2}{{\left(x+1\right)}^{2}}\right)dx}$
$=\displaystyle\int{{e}^{x}\dfrac{x-1}{x+1}}+\displaystyle\int{\dfrac{2{e}^{x}}{{\left(x+1\right)}^{2}}}$
$=\left(\dfrac{x-1}{x+1}\right)\times \displaystyle\int{{e}^{x}dx}-\displaystyle\int{\left[\dfrac{d}{dx}\left(\dfrac{x-1}{x+1}\right)\times \int{{e}^{x}dx}\right]dx}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$
$=\left(\dfrac{x-1}{x+1}\right){e}^{x}- \displaystyle\int{\left[\dfrac{\left(x+1\right)-\left(x-1\right)\times 1}{{\left(x+1\right)}^{2}}{e}^{x}\right]dx}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$
$=\left(\dfrac{x-1}{x+1}\right){e}^{x}-2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$
$={e}^{x}\left(\dfrac{x-1}{x+1}\right)+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Hard
Integration
$\displaystyle\int \dfrac{x^{2}+3}{x^{6}(x^{2}+1)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium

$\displaystyle \frac{x^{2}+2x+3}{x^{3}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}\Rightarrow A+B-C=$
• A. 6
• B. 3
• C. 1
• D.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate $\displaystyle \int \frac{\sec^{2}\:x}{\sqrt{\tan^{2}\:x+4}}dx.$
• A. $\displaystyle=\log\left | \tan\:x+\sqrt{4 \tan^{2}\:x+4} \right |+C$
• B. $\displaystyle=\log\left | \tan\:x-\sqrt{\tan^{2}\:x+4} \right |+C$
• C. $\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x-4} \right |+C$
• D. $\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x+4} \right |+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int_{0}^{\pi/2}\frac{1}{a+bcosx}dx=$, where $a>|b|$
• A. $\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a+b}{a-b}}$
• B. $\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}cot^{-l} \sqrt{\frac{a-b}{a+b}}$
• C. $\displaystyle \frac{\pi}{\sqrt{a^{2}-b^{2}}}$
• D. $\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a-b}{a+b}}$

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 $\displaystyle f(x)>0 \forall x \in R$