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

The integral of (x^3)/((x+3)^4) is ln|x+3|+9/(x+3)-(27)/(2(x+3)^2)+9/((x+3)^3)+C

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

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

ln ∣ x + 3 ∣ + \frac{9}{x + 3} - \frac{27}{2 ( x + 3 ) ^{2}} + \frac{9}{( x + 3 ) ^{3}} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3}}{( x + 3 ) ^{4}} : \frac{1}{x + 3} - \frac{9}{( x + 3 ) ^{2}} + \frac{27}{( x + 3 ) ^{3}} - \frac{27}{( x + 3 ) ^{4}}

= \int \frac{1}{x + 3} - \frac{9}{( x + 3 ) ^{2}} + \frac{27}{( x + 3 ) ^{3}} - \frac{27}{( x + 3 ) ^{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 + 3} d x - \int \frac{9}{( x + 3 ) ^{2}} d x + \int \frac{27}{( x + 3 ) ^{3}} d x - \int \frac{27}{( x + 3 ) ^{4}} d x

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

\int \frac{9}{( x + 3 ) ^{2}} d x = - \frac{9}{x + 3}

\int \frac{27}{( x + 3 ) ^{3}} d x = - \frac{27}{2 ( x + 3 ) ^{2}}

\int \frac{27}{( x + 3 ) ^{4}} d x = - \frac{9}{( x + 3 ) ^{3}}

= ln ∣ x + 3 ∣ - (- \frac{9}{x + 3}) - \frac{27}{2 ( x + 3 ) ^{2}} - (- \frac{9}{( x + 3 ) ^{3}})

Simplify

= ln ∣ x + 3 ∣ + \frac{9}{x + 3} - \frac{27}{2 ( x + 3 ) ^{2}} + \frac{9}{( x + 3 ) ^{3}}

Add a constant to the solution

= ln ∣ x + 3 ∣ + \frac{9}{x + 3} - \frac{27}{2 ( x + 3 ) ^{2}} + \frac{9}{( x + 3 ) ^{3}} + C

What is the integral of (x^3)/((x+3)^4) ?
The integral of (x^3)/((x+3)^4) is ln|x+3|+9/(x+3)-(27)/(2(x+3)^2)+9/((x+3)^3)+C