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

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

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integral of sqrt((x^3+1)/(x^3)) 1/(x^4)

\int \frac{x ^{3} + 1}{x ^{3}} \frac{1}{x ^{4}} d x

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

Solution steps

\int \frac{x ^{3} + 1}{x ^{3}} \frac{1}{x ^{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = \frac{x ^{3} + 1}{x ^{3}}

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

Simplify

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

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

Add a constant to the solution

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

f (y) = ln (y)

\int_{0}^{l n (π)} 2 x e^{x^{2}} sin (e^{x^{2}}) d x

\frac{d}{d x} (ln ((tan (x))^{3}))

\frac{d}{d x} (x^{2} - 5 x - 6)

\frac{d}{d u} ((u - u) (u + u))

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