What is the integral of (xe^{4x})/((4x+1)^2) ?

The integral of (xe^{4x})/((4x+1)^2) is (e^{4x})/(16(4x+1))+C

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integral of (xe^{4x})/((4x+1)^2)

\int \frac{x e ^{4 x}}{( 4 x + 1 ) ^{2}} d x

\frac{e ^{4 x}}{16 ( 4 x + 1 )} + C

Solution steps

\int \frac{x e ^{4 x}}{( 4 x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{e ^{u} u}{16 ( u + 1 ) ^{2}} d u

= \int \frac{e ^{u} u}{16 ( 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}{16} \cdot \int \frac{e ^{u} u}{( u + 1 ) ^{2}} d u

Apply Integration By Parts

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

= \frac{1}{16} (- \frac{e ^{u} u}{u + 1} - \int - e^{u} d u)

\int - e^{u} d u = - e^{u}

= \frac{1}{16} (- \frac{e ^{u} u}{u + 1} - (- e^{u}))

Substitute back

u = 4 x

= \frac{1}{16} (- \frac{e ^{4 x} \cdot 4 x}{4 x + 1} - (- e^{4 x}))

Simplify

\frac{1}{16} (- \frac{e ^{4 x} \cdot 4 x}{4 x + 1} - (- e^{4 x})) : \frac{e ^{4 x}}{16 ( 4 x + 1 )}

= \frac{e ^{4 x}}{16 ( 4 x + 1 )}

Add a constant to the solution

= \frac{e ^{4 x}}{16 ( 4 x + 1 )} + C

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What is the integral of (xe^{4x})/((4x+1)^2) ?
The integral of (xe^{4x})/((4x+1)^2) is (e^{4x})/(16(4x+1))+C