Mathematics

# Solve:$\int {\frac{{\sqrt x }}{{1 + x}}} dx$

##### SOLUTION
$I=\displaystyle \int \dfrac {\sqrt {a}}{1+x}dx$
put $x=t^2$
$dx =2t\ dt$
$dx=2\sqrt {x}dt$
$\dfrac {dx}{\sqrt x}=2dt \Rightarrow \dfrac {\sqrt {x}dx}{x}=2dt$
$\Rightarrow \ \sqrt x\ dx=2t^2\ dt$
$I=\displaystyle \int \dfrac {2t^2dt}{1+t^2}=\displaystyle \int \dfrac {2(t^2+1)}{1+t^2}dt -2\displaystyle \int \dfrac {dt}{1+t^2}$
$=\displaystyle \int 2dt -2\displaystyle \int \dfrac {dt}{1+t^2}$
$I=2t-2\tan^{-1}(t)+c$
$I=2\sqrt {x}-2\tan^{-1}(\sqrt {x})+c$

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

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle \frac{(x-1)^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$, then $A=..., B=..., C=...$
• A. $A=1,B=0,C=2$
• B. $A=0,B=1,C=-2$
• C. $A=0,B=1,C=2$
• D. $A=1,B=0,C=-2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$ then I is equal to
• A. $\displaystyle \pi$
• B. $\displaystyle \pi -\frac{1}{\pi }$
• C. $\displaystyle \pi +\frac{1}{\pi }$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int\frac{\cos x}{\sin^{2}x+4\sin x+5}dx$ is equal to
• A. $\sin^{-1} ( \sin x+2) +c$
• B. $\cot^{-1} (\sin x+2) +c$
• C. $\sec^{-1} (sin x+2) +c$
• D. $\tan^{-1} ( \sin x+2) +c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\int { { e }^{ \tan { \theta } } } \left( \sec { \theta } -\sin { \theta } \right) d\theta$ is equal to ?
• A. $-{ e }^{ \tan { \theta } }\sin { \theta } +C$
• B. ${ e }^{ \tan { \theta } }\sin { \theta } +C$
• C. ${ e }^{ \tan { \theta } }\sec { \theta } +C$
• D. ${ e }^{ \tan { \theta } }\cos { \theta } +C$

1 Verified Answer | Published on 17th 09, 2020

Q5 Subjective Medium
Solve : $\displaystyle \int \, \dfrac{dx}{x - \sqrt{x}}$