What is the integral of ((ln(2x))/((x+1)^3)) ?

The integral of ((ln(2x))/((x+1)^3)) is -(ln(2x))/(2(x+1)^2)+1/2 (ln|x|-ln|x+1|+1/(x+1))+C

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integral of ((ln(2x))/((x+1)^3))

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

- \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} + \frac{1}{2} (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) + C

Solution steps

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

Apply Integration By Parts

: - \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} - \int - \frac{1}{2 x ( x + 1 ) ^{2}} d x

= - \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} - \int - \frac{1}{2 x ( x + 1 ) ^{2}} d x

\int - \frac{1}{2 x ( x + 1 ) ^{2}} d x = - \frac{1}{2} (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

= - \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} - (- \frac{1}{2} (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}))

Simplify

= - \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} + \frac{1}{2} (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Add a constant to the solution

= - \frac{ln ( 2 x )}{2 ( x + 1 ) ^{2}} + \frac{1}{2} (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) + C

\frac{d}{d x} (e^{x} - x^{2})

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

\frac{d x}{d t} + 4 t^{5} x^{7} + \frac{x}{t} = 0

y^{^{''}} - \frac{2}{x ^{2}} y = 0, y (- 1) = - 1, y^{^{'}} (- 1) = - 4

\int (9 t^{2} + \frac{1}{t ^{3}}) d t

What is the integral of ((ln(2x))/((x+1)^3)) ?
The integral of ((ln(2x))/((x+1)^3)) is -(ln(2x))/(2(x+1)^2)+1/2 (ln|x|-ln|x+1|+1/(x+1))+C