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

The integral of (x^2-1/(x^2))^3x is (x^8)/8-(3x^4)/4+3ln|x|+1/(4x^4)+C

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

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

\frac{x ^{8}}{8} - \frac{3 x ^{4}}{4} + 3 ln ∣ x ∣ + \frac{1}{4 x ^{4}} + C

Solution steps

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

Expand

(x^{2} - \frac{1}{x ^{2}})^{3} x : x^{7} - 3 x^{3} + \frac{3}{x} - \frac{1}{x ^{5}}

= \int x^{7} - 3 x^{3} + \frac{3}{x} - \frac{1}{x ^{5}} 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 x^{7} d x - \int 3 x^{3} d x + \int \frac{3}{x} d x - \int \frac{1}{x ^{5}} d x

\int x^{7} d x = \frac{x ^{8}}{8}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

= \frac{x ^{8}}{8} - \frac{3 x ^{4}}{4} + 3 ln ∣ x ∣ - (- \frac{1}{4 x ^{4}})

Simplify

= \frac{x ^{8}}{8} - \frac{3 x ^{4}}{4} + 3 ln ∣ x ∣ + \frac{1}{4 x ^{4}}

Add a constant to the solution

= \frac{x ^{8}}{8} - \frac{3 x ^{4}}{4} + 3 ln ∣ x ∣ + \frac{1}{4 x ^{4}} + C

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What is the integral of (x^2-1/(x^2))^3x ?
The integral of (x^2-1/(x^2))^3x is (x^8)/8-(3x^4)/4+3ln|x|+1/(4x^4)+C