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

\int \frac{2}{x ( x ^{2} + 4 ) ^{2}} d x

y y^{^{'}} - 8 e^{x} = 0

ln (2 x + 5)

sin (x^{2})

\int e^{2 t} t d t

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