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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}))

ln (1 + 3 x)

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{4 x})

y^{^{'}} = 2 + 3 x + 8 y + 12 x y

- 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

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