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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int u d u

Apply the Power Rule

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

Substitute back

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

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

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

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

Add a constant to the solution

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

\int sin^{6} (5 x) d x

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

n = 1 \sum \infty (\frac{2}{3})^{n + 1}

d er i v a t i v e f (x) = sin (x^{2} + 2 x + 3)

a re a y = 2 x^{2} - 1, [0, 1]

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