Mathematics

# Evaluate:  $\int {\frac {e^{2x}}{1+e^x}}dx$

##### SOLUTION
$l = (1-e^{x})-log|e^{x}+1|+c$
$= (t-logt)+c$
$= (e^{x}+1)-log(e^{x}+1)+c$
$= e^{x}-log(e^{x}+1)+1+c$
$= e^{x}-log(e^{x}+1)+c$

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

#### Realted Questions

Q1 TRUE/FALSE Medium
State whether the given statement is true or false
the value of the integral $\displaystyle \int_{0}^{2a}\frac{f\left ( x \right )}{f\left ( x \right )+f\left ( 2a-x \right )}dx$ is equal to a.
• A. False
• B. True

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\int \sin x \sin(\cos x) \ dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle h\left( x \right) =\int _{ 0 }^{ x }{ { \sin }^{ 4 }t } dt$, then $\displaystyle h\left( x+\pi \right)$ equals
• A. $\displaystyle \frac { h\left( x \right) }{ h\left( \pi \right) }$
• B. $\displaystyle h\left( x \right) h\left( \pi \right)$
• C. $\displaystyle h\left( x \right) -h\left( \pi \right)$
• D. $\displaystyle h\left( x \right) +h\left( \pi \right)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Let $\displaystyle f\left ( x \right )=min\left \{ x,1-x \right \}$ for all $x \epsilon R$. Then the value of $\displaystyle \int_{0}^{2}f\left ( x \right )dx$ is
• A. $\displaystyle \frac{1}{4}$
• B. $2$
• C. $0$
• D. $\displaystyle -\frac{1}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.