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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = x^{3} + x^{2}

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

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

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

Add a constant to the solution

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

\int (\frac{5}{x} + \frac{15}{x ^{6}}) d x

t - 4

(x + y) \cdot y^{^{'}} = 2 \cdot x + 2 \cdot y - 3

\int \frac{e ^{x - 1}}{x} d x

35 x - x x

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