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

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

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integral of (x^2)/(sqrt(x+1))

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \int u^{\frac{3}{2}} - 2 u + \frac{1}{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{3}{2}} d u - \int 2 u d u + \int \frac{1}{u} d u

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

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

\int \frac{1}{u} d u = 2 u

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

x \to 0 - lim (1 + e^{\frac{1}{x}})

3 t^{2} + 15 t + 15

y = (5 x^{2} - 1)^{- 3}

\int_{- \infty}^{0} x e^{8 x} d x

2 x^{2} - 4 x

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