What is the integral of (6-x)sqrt(x+1) ?

The integral of (6-x)sqrt(x+1) is -2/5 (x+1)^{5/2}+14/3 (x+1)^{3/2}+C

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integral of (6-x)sqrt(x+1)

\int (6 - x) x + 1 d x

- \frac{2}{5} (x + 1)^{\frac{5}{2}} + \frac{14}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int (6 - x) x + 1 d x

Expand

(6 - x) x + 1 : 6 x + 1 - x x + 1

= \int 6 x + 1 - x x + 1 d x

Apply the Sum Rule:

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

= \int 6 x + 1 d x - \int x x + 1 d x

\int 6 x + 1 d x = 4 (x + 1)^{\frac{3}{2}}

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

= 4 (x + 1)^{\frac{3}{2}} - (\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}})

Simplify

4 (x + 1)^{\frac{3}{2}} - (\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}}) : - \frac{2}{5} (x + 1)^{\frac{5}{2}} + \frac{14}{3} (x + 1)^{\frac{3}{2}}

= - \frac{2}{5} (x + 1)^{\frac{5}{2}} + \frac{14}{3} (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{5} (x + 1)^{\frac{5}{2}} + \frac{14}{3} (x + 1)^{\frac{3}{2}} + C

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What is the integral of (6-x)sqrt(x+1) ?
The integral of (6-x)sqrt(x+1) is -2/5 (x+1)^{5/2}+14/3 (x+1)^{3/2}+C