What is the integral of x^2sqrt(6+x) ?

The integral of x^2sqrt(6+x) is 2/7 (6+x)^{7/2}-24/5 (6+x)^{5/2}+24(6+x)^{3/2}+C

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integral of x^2sqrt(6+x)

\int x^{2} 6 + x d x

\frac{2}{7} (6 + x)^{\frac{7}{2}} - \frac{24}{5} (6 + x)^{\frac{5}{2}} + 24 (6 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 6 + x d x

Apply u-substitution

: \int (u - 6)^{2} u d u

= \int (u - 6)^{2} u d u

Expand

(u - 6)^{2} u : u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 u

= \int u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 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 12 u^{\frac{3}{2}} d u + \int 36 u d u

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

\int 12 u^{\frac{3}{2}} d u = \frac{24}{5} u^{\frac{5}{2}}

\int 36 u d u = 24 u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{24}{5} u^{\frac{5}{2}} + 24 u^{\frac{3}{2}}

Substitute back

u = 6 + x

= \frac{2}{7} (6 + x)^{\frac{7}{2}} - \frac{24}{5} (6 + x)^{\frac{5}{2}} + 24 (6 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (6 + x)^{\frac{7}{2}} - \frac{24}{5} (6 + x)^{\frac{5}{2}} + 24 (6 + x)^{\frac{3}{2}} + C

\int_{- 3}^{- 2} \frac{1}{x} d x

\int 0.5 e^{- 0.5 x} d x

\frac{d}{d x} (4 π)

5 s

\frac{c e ^{t}}{1 + c e ^{t}}

What is the integral of x^2sqrt(6+x) ?
The integral of x^2sqrt(6+x) is 2/7 (6+x)^{7/2}-24/5 (6+x)^{5/2}+24(6+x)^{3/2}+C