What is the integral of 1/(x^3)(1+1/(x^2))^{5/3} ?

The integral of 1/(x^3)(1+1/(x^2))^{5/3} is -(3(x^2+1)^{8/3})/(16x^{16/3)}+C

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integral of 1/(x^3)(1+1/(x^2))^{5/3}

\int \frac{1}{x ^{3}} (1 + \frac{1}{x ^{2}})^{\frac{5}{3}} d x

- \frac{3 ( x ^{2} + 1 ) ^{\frac{8}{3}}}{16 x ^{\frac{16}{3}}} + C

Solution steps

\int \frac{1}{x ^{3}} (1 + \frac{1}{x ^{2}})^{\frac{5}{3}} d x

Apply u-substitution

: \int - \frac{u ^{\frac{5}{3}}}{2} d u

= \int - \frac{u ^{\frac{5}{3}}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int u^{\frac{5}{3}} d u

Apply the Power Rule

: \frac{3}{8} u^{\frac{8}{3}}

= - \frac{1}{2} \cdot \frac{3}{8} u^{\frac{8}{3}}

Substitute back

u = 1 + \frac{1}{x ^{2}}

= - \frac{1}{2} \cdot \frac{3}{8} (1 + \frac{1}{x ^{2}})^{\frac{8}{3}}

Simplify

- \frac{1}{2} \cdot \frac{3}{8} (1 + \frac{1}{x ^{2}})^{\frac{8}{3}} : - \frac{3 ( x ^{2} + 1 ) ^{\frac{8}{3}}}{16 x ^{\frac{16}{3}}}

= - \frac{3 ( x ^{2} + 1 ) ^{\frac{8}{3}}}{16 x ^{\frac{16}{3}}}

Add a constant to the solution

= - \frac{3 ( x ^{2} + 1 ) ^{\frac{8}{3}}}{16 x ^{\frac{16}{3}}} + C

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What is the integral of 1/(x^3)(1+1/(x^2))^{5/3} ?
The integral of 1/(x^3)(1+1/(x^2))^{5/3} is -(3(x^2+1)^{8/3})/(16x^{16/3)}+C