What is the integral of (e^{2x})/((2x+1)^2) ?

The integral of (e^{2x})/((2x+1)^2) is 1/2 (-(e^{2x})/(2x+1)+1/e \Ei(2x+1))+C

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integral of (e^{2x})/((2x+1)^2)

\int \frac{e ^{2 x}}{( 2 x + 1 ) ^{2}} d x

\frac{1}{2} (- \frac{e ^{2 x}}{2 x + 1} + \frac{1}{e} Ei (2 x + 1)) + C

Solution steps

\int \frac{e ^{2 x}}{( 2 x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{2 ( u + 1 ) ^{2}} d u

= \int \frac{e ^{u}}{2 ( u + 1 ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{e ^{u}}{( u + 1 ) ^{2}} d u

Apply Integration By Parts

: - \frac{e ^{u}}{u + 1} - \int - \frac{e ^{u}}{u + 1} d u

= \frac{1}{2} (- \frac{e ^{u}}{u + 1} - \int - \frac{e ^{u}}{u + 1} d u)

\int - \frac{e ^{u}}{u + 1} d u = - \frac{1}{e} Ei (u + 1)

= \frac{1}{2} (- \frac{e ^{u}}{u + 1} - (- \frac{1}{e} Ei (u + 1)))

Substitute back

u = 2 x

= \frac{1}{2} (- \frac{e ^{2 x}}{2 x + 1} - (- \frac{1}{e} Ei (2 x + 1)))

Simplify

= \frac{1}{2} (- \frac{e ^{2 x}}{2 x + 1} + \frac{1}{e} Ei (2 x + 1))

Add a constant to the solution

= \frac{1}{2} (- \frac{e ^{2 x}}{2 x + 1} + \frac{1}{e} Ei (2 x + 1)) + C

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What is the integral of (e^{2x})/((2x+1)^2) ?
The integral of (e^{2x})/((2x+1)^2) is 1/2 (-(e^{2x})/(2x+1)+1/e \Ei(2x+1))+C