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

The integral of (1/(x^3)+3)^2 is -1/(5x^5)-3/(x^2)+9x+C

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

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

- \frac{1}{5 x ^{5}} - \frac{3}{x ^{2}} + 9 x + C

Solution steps

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

Expand

(\frac{1}{x ^{3}} + 3)^{2} : \frac{1}{x ^{6}} + \frac{6}{x ^{3}} + 9

= \int \frac{1}{x ^{6}} + \frac{6}{x ^{3}} + 9 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 ^{6}} d x + \int \frac{6}{x ^{3}} d x + \int 9 d x

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

\int \frac{6}{x ^{3}} d x = - \frac{3}{x ^{2}}

\int 9 d x = 9 x

= - \frac{1}{5 x ^{5}} - \frac{3}{x ^{2}} + 9 x

Add a constant to the solution

= - \frac{1}{5 x ^{5}} - \frac{3}{x ^{2}} + 9 x + C

What is the integral of (1/(x^3)+3)^2 ?
The integral of (1/(x^3)+3)^2 is -1/(5x^5)-3/(x^2)+9x+C