Mathematics

# $\displaystyle \int { \frac { \cos { x } +x\sin { x } }{ x\left( x+\cos { x } \right) } dx }$ is equal to

$\displaystyle \log { \left| \frac { x }{ x+\cos { x } } \right| +c }$

##### SOLUTION
$\displaystyle \int { \frac { \cos { x } +x\sin { x } }{ x\left( x+\cos { x } \right) } dx } =\int { \frac { \left( x+\cos { x } \right) -x\left( 1-\sin { x } \right) }{ x\left( x+\cos { x } \right) } dx }$

$\displaystyle =\int { \left( \frac { 1 }{ x } -\frac { 1-\sin { x } }{ x+\cos { x } } \right) dx } =\log { \left| x \right| } -\log { \left| x+\cos { x } \right| } +c$

$\displaystyle =\log { \left| \frac { x }{ x+\cos { x } } \right| +c }$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Subjective Medium
Integrate : $\displaystyle \int e^x \cdot \frac {x}{(1+x)^2}dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Integrate $\int {({{\sin }^{ - 1}}} x{)^2}dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I_1=\int_{0}^{\frac{\pi}{2}}e^{-x} sin 4x dx$  and $\displaystyle I_2=\int_0^{2\pi}e^{-x} sin 4x dx$ and  $\displaystyle I_2=\lambda I_1$, then $\displaystyle \lambda$ is equal to
• A. $\displaystyle \frac{e^{2\pi}-1}{e^{\pi}-1}$
• B. $\displaystyle \frac{e^{\pi}-1}{1-e^{2\pi}}$
• C. $\displaystyle \frac{1-e^{2\pi}}{1-e^{\frac{\pi}{2}}}$
• D. $\displaystyle \frac{1-e^{-2\pi}}{1-e^{-\frac{\pi}{2}}}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium

Evaluate the following definite integral:

$\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Subjective Medium
Evaluate:
$\displaystyle\int\limits_{-a}^{a}x^3\sqrt{x^2-a^2}\ dx$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020