What is the integral of (x^2+2x-10)^{2/3}(5x+5) ?

The integral of (x^2+2x-10)^{2/3}(5x+5) is 3/2 (x^2+2x-10)^{5/3}+C

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integral of (x^2+2x-10)^{2/3}(5x+5)

\int (x^{2} + 2 x - 10)^{\frac{2}{3}} (5 x + 5) d x

\frac{3}{2} (x^{2} + 2 x - 10)^{\frac{5}{3}} + C

Solution steps

\int (x^{2} + 2 x - 10)^{\frac{2}{3}} (5 x + 5) d x

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{5}{2} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = x^{2} + 2 x - 10

= \frac{5}{2} \cdot \frac{3}{5} (x^{2} + 2 x - 10)^{\frac{5}{3}}

Simplify

\frac{5}{2} \cdot \frac{3}{5} (x^{2} + 2 x - 10)^{\frac{5}{3}} : \frac{3}{2} (x^{2} + 2 x - 10)^{\frac{5}{3}}

= \frac{3}{2} (x^{2} + 2 x - 10)^{\frac{5}{3}}

Add a constant to the solution

= \frac{3}{2} (x^{2} + 2 x - 10)^{\frac{5}{3}} + C

l a pl a ce t r an s f or m e^{- 2 t} - 3 cos (2 t)

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

\frac{d}{d x} (1 + tan (x))

d er i v a t i v e f (X) = 4 X^{3} - 1

\int t^{n} d t

What is the integral of (x^2+2x-10)^{2/3}(5x+5) ?
The integral of (x^2+2x-10)^{2/3}(5x+5) is 3/2 (x^2+2x-10)^{5/3}+C