What is the integral of (9/(x^4)+x^2) ?

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

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integral of (9/(x^4)+x^2)

\int (\frac{9}{x ^{4}} + x^{2}) d x

- \frac{3}{x ^{3}} + \frac{x ^{3}}{3} + C

Solution steps

\int \frac{9}{x ^{4}} + x^{2} 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{9}{x ^{4}} d x + \int x^{2} d x

\int \frac{9}{x ^{4}} d x = - \frac{3}{x ^{3}}

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

= - \frac{3}{x ^{3}} + \frac{x ^{3}}{3}

Add a constant to the solution

= - \frac{3}{x ^{3}} + \frac{x ^{3}}{3} + C

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