What is the integral of 3/(x^4)sqrt(8-1/(x^3)) ?

The integral of 3/(x^4)sqrt(8-1/(x^3)) is (2(8x^3-1)^{3/2})/(3x^{9/2)}+C

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integral of 3/(x^4)sqrt(8-1/(x^3))

\int \frac{3}{x ^{4}} 8 - \frac{1}{x ^{3}} d x

\frac{2 ( 8 x ^{3} - 1 ) ^{\frac{3}{2}}}{3 x ^{\frac{9}{2}}} + C

Solution steps

\int \frac{3}{x ^{4}} 8 - \frac{1}{x ^{3}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{4}} 8 - \frac{1}{x ^{3}} d x

Apply u-substitution

: \int \frac{u}{3} d u

= 3 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 8 - \frac{1}{x ^{3}}

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

Simplify

3 \cdot \frac{1}{3} \cdot \frac{2}{3} (8 - \frac{1}{x ^{3}})^{\frac{3}{2}} : \frac{2 ( 8 x ^{3} - 1 ) ^{\frac{3}{2}}}{3 x ^{\frac{9}{2}}}

= \frac{2 ( 8 x ^{3} - 1 ) ^{\frac{3}{2}}}{3 x ^{\frac{9}{2}}}

Add a constant to the solution

= \frac{2 ( 8 x ^{3} - 1 ) ^{\frac{3}{2}}}{3 x ^{\frac{9}{2}}} + C

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

\int 2 cos^{2} (x) tan^{2} (x) d x

(ln (x))^{4}

\frac{d ^{2}}{d x ^{2}} (cos (2 x))

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

What is the integral of 3/(x^4)sqrt(8-1/(x^3)) ?
The integral of 3/(x^4)sqrt(8-1/(x^3)) is (2(8x^3-1)^{3/2})/(3x^{9/2)}+C