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

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

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integral of sqrt(x^3+x^2)

\int x^{3} + x^{2} d x

\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} + x^{2} d x

Factor

x^{3} + x^{2} : x^{2} x + 1

= \int x^{2} x + 1 d x

x^{2} = (x),

assuming

x \geq 0

= \int x x + 1 d x

Apply u-substitution

= \int (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

u = x + 1

= \frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

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