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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Expand

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

= 2 \cdot \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

= 2 (\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

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

Substitute back

u = x + 1

= 2 (\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

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

\frac{d}{d t} (t^{2} + 1)

\frac{x}{x + 2}

5 y^{^{'}} + y = 0

\int_{1}^{7} \frac{x ^{2} - 1}{x} d x

\frac{d}{d t} (t - 1)

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