What is the integral of x^2sqrt(x-7) ?

The integral of x^2sqrt(x-7) is 2/7 (x-7)^{7/2}+28/5 (x-7)^{5/2}+98/3 (x-7)^{3/2}+C

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integral of x^2sqrt(x-7)

\int x^{2} x - 7 d x

\frac{2}{7} (x - 7)^{\frac{7}{2}} + \frac{28}{5} (x - 7)^{\frac{5}{2}} + \frac{98}{3} (x - 7)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x - 7 d x

Apply u-substitution

: \int (u + 7)^{2} u d u

= \int (u + 7)^{2} u d u

Expand

(u + 7)^{2} u : u^{\frac{5}{2}} + 14 u^{\frac{3}{2}} + 49 u

= \int u^{\frac{5}{2}} + 14 u^{\frac{3}{2}} + 49 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 14 u^{\frac{3}{2}} d u + \int 49 u d u

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

\int 14 u^{\frac{3}{2}} d u = \frac{28}{5} u^{\frac{5}{2}}

\int 49 u d u = \frac{98}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} + \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}}

Substitute back

u = x - 7

= \frac{2}{7} (x - 7)^{\frac{7}{2}} + \frac{28}{5} (x - 7)^{\frac{5}{2}} + \frac{98}{3} (x - 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x - 7)^{\frac{7}{2}} + \frac{28}{5} (x - 7)^{\frac{5}{2}} + \frac{98}{3} (x - 7)^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(x-7) ?
The integral of x^2sqrt(x-7) is 2/7 (x-7)^{7/2}+28/5 (x-7)^{5/2}+98/3 (x-7)^{3/2}+C