What is the integral of x^2sqrt(x+4) ?

The integral of x^2sqrt(x+4) is 2/7 (x+4)^{7/2}-16/5 (x+4)^{5/2}+32/3 (x+4)^{3/2}+C

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integral of x^2sqrt(x+4)

\int x^{2} x + 4 d x

\frac{2}{7} (x + 4)^{\frac{7}{2}} - \frac{16}{5} (x + 4)^{\frac{5}{2}} + \frac{32}{3} (x + 4)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x + 4 d x

Apply u-substitution

: \int (u - 4)^{2} u d u

= \int (u - 4)^{2} u d u

Expand

(u - 4)^{2} u : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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{5}{2}} d u - \int 8 u^{\frac{3}{2}} d u + \int 16 u d u

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

\int 16 u d u = \frac{32}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}}

Substitute back

u = x + 4

= \frac{2}{7} (x + 4)^{\frac{7}{2}} - \frac{16}{5} (x + 4)^{\frac{5}{2}} + \frac{32}{3} (x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x + 4)^{\frac{7}{2}} - \frac{16}{5} (x + 4)^{\frac{5}{2}} + \frac{32}{3} (x + 4)^{\frac{3}{2}} + C

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

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

x \to 1 + lim (x^{2} - 4 x + 3)

\frac{\partial}{\partial x} (4 e^{\frac{x ^{2}}{y}} - cos (x y^{3}))

x \to 0 lim (4 x + x)

What is the integral of x^2sqrt(x+4) ?
The integral of x^2sqrt(x+4) is 2/7 (x+4)^{7/2}-16/5 (x+4)^{5/2}+32/3 (x+4)^{3/2}+C