Mathematics

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

$e^{x}\dfrac{1}{x+1}$

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

#### Realted Questions

Q1 Single Correct Hard
Evaluate $\displaystyle\int{\frac{dx}{1+\sqrt{x^2+2x+2}}}$.
• A. $\displaystyle I=\ln{(x+1-\sqrt{x^2+2x+2})}+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$
• B. $\displaystyle I=\ln{(x-2-\sqrt{x^2-2x-4})}+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$
• C. None of these
• D. $\displaystyle I=\ln{(x+1+\sqrt{x^2+2x+2})}+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int^2_1x^{-5}dx$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The following integral $\displaystyle \int_{\pi/4}^{\pi/2} (2 cosec x)^{17}dx$ is equal to
• A. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{17} du$
• B. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{17} du$
• C. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du$
• D. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate: $\displaystyle\int \tan^{-1}\sqrt{x}dx$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020