What is the integral of x^2sqrt(x+1) ?

The integral of x^2sqrt(x+1) is 2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2}+C

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integral of x^2sqrt(x+1)

\int x^{2} x + 1 d x

\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x + 1 d x

Apply u-substitution

: \int (u - 1)^{2} u d u

= \int (u - 1)^{2} u d u

Expand

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

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

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

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x + 1

= \frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (sin (x) + x)

\int \frac{2 x}{1 - 4 x ^{4}} d x

\int \frac{z ^{2}}{z ^{3} + 1} d z

- 9 (5 x^{2} + 4)^{- 6}

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

What is the integral of x^2sqrt(x+1) ?
The integral of x^2sqrt(x+1) is 2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2}+C