Mathematics

# Evaluate: $\displaystyle \int \dfrac x{1-x^2} dx$

##### SOLUTION
$I=\displaystyle \int \dfrac{x}{1-x^2}dx$

$t=1-x^2\implies dt=-2x dx$

$=\displaystyle \int -\dfrac 12\dfrac 1t dt$

$=-\dfrac 12\log t$

$=-\dfrac 12\log (1-x^2)+c$

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

#### Realted Questions

Q1 One Word Medium
$\displaystyle \int {\dfrac{7^{2x+3}\sin^2 2x+ \cos^22x}{\sin^22x}}=\dfrac{7^{2x+3}}{2\log 7}-\dfrac{(\cot x+x)}{b}$.Find $b$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate $\displaystyle \int 3^{x}\cos4xdx=$
• A. $\displaystyle \dfrac{3^{x}}{9+(\log 4)^{2}}[\cos4x+(\log 3)\cdot \sin4x]+c$
• B. $\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+(\log 4)^{2}}[(\log 3)\sin4x+(\log 4)\cos3x]+c$
• C. $\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\sin4x+(\log 4)\cos3x]+c$
• D. $\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\cos4x+4\sin4x]+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Integrate $\displaystyle\int^1_03^xdx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\int \sqrt{1 +\sin2x} dx$ is
• A. $\sin 2x + c$
• B. $\cos x + \sin x + c$
• C. $\cos 2x + c$
• D. $\sin x - \cos x + c$

Solve $\int {\sqrt {\dfrac{{1 + x}}{{1 - x}}} }$