What is the integral of (x^3)/((x+1)^4) ?

The integral of (x^3)/((x+1)^4) is ln|x+1|+3/(x+1)-3/(2(x+1)^2)+1/(3(x+1)^3)+C

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integral of (x^3)/((x+1)^4)

\int \frac{x ^{3}}{( x + 1 ) ^{4}} d x

ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + C

Solution steps

\int \frac{x ^{3}}{( x + 1 ) ^{4}} d x

Take the partial fraction of

\frac{x ^{3}}{( x + 1 ) ^{4}} : \frac{1}{x + 1} - \frac{3}{( x + 1 ) ^{2}} + \frac{3}{( x + 1 ) ^{3}} - \frac{1}{( x + 1 ) ^{4}}

= \int \frac{1}{x + 1} - \frac{3}{( x + 1 ) ^{2}} + \frac{3}{( x + 1 ) ^{3}} - \frac{1}{( x + 1 ) ^{4}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x + 1} d x - \int \frac{3}{( x + 1 ) ^{2}} d x + \int \frac{3}{( x + 1 ) ^{3}} d x - \int \frac{1}{( x + 1 ) ^{4}} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

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

\int \frac{1}{( x + 1 ) ^{4}} d x = - \frac{1}{3 ( x + 1 ) ^{3}}

= ln ∣ x + 1 ∣ - (- \frac{3}{x + 1}) - \frac{3}{2 ( x + 1 ) ^{2}} - (- \frac{1}{3 ( x + 1 ) ^{3}})

Simplify

= ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + C

What is the integral of (x^3)/((x+1)^4) ?
The integral of (x^3)/((x+1)^4) is ln|x+1|+3/(x+1)-3/(2(x+1)^2)+1/(3(x+1)^3)+C