Mathematics

# Evaluate: $\displaystyle \int \frac{1}{(1-x)\sqrt{1+x^2}}dx=$

$\dfrac{\log(|\sqrt{x^2+1}+(\sqrt2+1)x-1|)-log(|\sqrt{x^2+1}+(1-\sqrt2)x-1|)}{\sqrt2}+c$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
The integral of $\displaystyle \frac{\cos ^{3}x}{\sin ^{2}x+\sin x}$ gives $f(x)$ with const . of integration as zero. find $f( \pi/2 )$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Solve $\int {\dfrac { 1-{ x }^{ 7 } }{ x\left( 1+{ x }^{ 7 } \right) } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate $\displaystyle\int{\frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx}$.
• A. $\displaystyle\frac{{(x+\sqrt{1+x^2})}^{14}}{14}+C$
• B. $\displaystyle\frac{{(x+\sqrt{1+x^2})}^{16}}{16}+C$
• C. $\displaystyle\frac{{(x+\sqrt{1+x^2})}^{17}}{17}+C$
• D. $\displaystyle\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int{\dfrac{1}{\cos^{2}x(1-\tan x)^{2}}dx}$=
• A. $\dfrac{1}{1-\tan x}+c$
• B. $-\dfrac{1}{3}\dfrac{1}{(1-\tan x)^{3}}+c$
• C. $none\ of\ these$
• D. $\dfrac{1}{\tan x-1}+c$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$