Mathematics

If $\displaystyle \int \frac{\cos x- \sin x+1-x}{e^x+\sin x +x}dx-\ln \left ( f(x) \right )+g(x)+C$ where $C$ is the constant of integration and $f(x)$ is positive , then $f(x)+g(x)$ has the value equal to

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

SOLUTION
$\displaystyle I=\int { \frac { \cos { x-\sin { x+1-x } } }{ { e }^{ x }+\sin { x } +x } } dx$

$\displaystyle =\int { \frac { { e }^{ x }+\cos { x } +1-{ e }^{ x }-\sin { x } -x }{ { e }^{ x }+\sin { x } +x } } dx$

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

$=\log { \left( { e }^{ 2 }+\sin { x+x } \right) -x }$
$\displaystyle \because \frac { d }{ dx } \left( { e }^{ x }+\sin { x } +x \right) ={ e }^{ x }+\cos { x } +1$

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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 Medium
$\int _{ 0 }^{ \pi }{ |cosx{ | }^{ 3 }dx }$ is equal to
• A. $\cfrac { 2 }{ 3 }$
• B. $0$
• C. $\cfrac { -8 }{ 3 }$
• D. $\cfrac { 4 }{ 3 }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle\int \dfrac { d x } { e ^ { x } + 1 - 2 e ^ { - x } } =$
• A. $\log \left| e ^ { x } - 1 \right| - \log \left| e ^ { x } + 2 \right| + c$
• B. $\frac { 1 } { 2 } \log \left| e ^ { x } - 1 \right| - \frac { 1 } { 3 } \log \left| e ^ { x } + 2 \right| + c$
• C. $\frac { 1 } { 3 } \log \left| e ^ { x } - 1 \right| + \frac { 1 } { 3 } \log \left| e ^ { x } + 2 \right| + c$
• D. $\frac { 1 } { 3 } \log \left| e ^ { x } - 1 \right| - \frac { 1 } { 3 } \log \left| e ^ { x } + 2 \right| + c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Let f(x) be a continuous function such that $f(a-x)+f(x)=0$ for all $x\in [0, a]$. Then $\displaystyle\int^a_0\dfrac{dx}{1+e^{f(x)}}$ is equal to?
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• D. a

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate
$\displaystyle\int { \dfrac { \cos { \sqrt { x } } }{ \sqrt { x } } } dx$

Prove that $\displaystyle\int_0^{\pi/2}$ $ln(\sin x)dx=\displaystyle\int_0^{\pi/2}ln(cos x)dx=\int_0^{\pi/2}\,\,ln(sin2x)dx=-\dfrac{\pi}{2}.ln 2$.