What is the integral of (xe^{3x})/((1+3x)^2) ?

The integral of (xe^{3x})/((1+3x)^2) is (e^{3x})/(9(1+3x))+C

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integral of (xe^{3x})/((1+3x)^2)

\int \frac{x e ^{3 x}}{( 1 + 3 x ) ^{2}} d x

\frac{e ^{3 x}}{9 ( 1 + 3 x )} + C

Solution steps

\int \frac{x e ^{3 x}}{( 1 + 3 x ) ^{2}} d x

Apply u-substitution

: \int \frac{e ^{u} u}{9 ( 1 + u ) ^{2}} d u

= \int \frac{e ^{u} u}{9 ( 1 + u ) ^{2}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{e ^{u} u}{( 1 + u ) ^{2}} d u

Apply Integration By Parts

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

= \frac{1}{9} (- \frac{e ^{u} u}{1 + u} - \int - e^{u} d u)

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

= \frac{1}{9} (- \frac{e ^{u} u}{1 + u} - (- e^{u}))

Substitute back

u = 3 x

= \frac{1}{9} (- \frac{e ^{3 x} \cdot 3 x}{1 + 3 x} - (- e^{3 x}))

Simplify

\frac{1}{9} (- \frac{e ^{3 x} \cdot 3 x}{1 + 3 x} - (- e^{3 x})) : \frac{e ^{3 x}}{9 ( 1 + 3 x )}

= \frac{e ^{3 x}}{9 ( 1 + 3 x )}

Add a constant to the solution

= \frac{e ^{3 x}}{9 ( 1 + 3 x )} + C

\int 3 tan (x) sec^{2} (x) d x

f (x) = 800 - 800 x^{- 1}

y = \frac{2 x + 5}{x}

\int sec^{4} (x) d x

f (x) = 60 + 20 x - 4.5 x^{2}, at x = 2

What is the integral of (xe^{3x})/((1+3x)^2) ?
The integral of (xe^{3x})/((1+3x)^2) is (e^{3x})/(9(1+3x))+C