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

The integral of (x^2)/(sqrt(x)+1) is 2/5 x^{5/2}-(x^2)/2+2/3 x^{3/2}-x+2sqrt(x)-2ln|sqrt(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^{\frac{5}{2}} - \frac{x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} - x + 2 x - 2 ln x + 1 + C

Solution steps

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

Apply u-substitution

= \int \frac{2 u ^{5}}{u + 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{5}}{u + 1} d u

Long division

\frac{u ^{5}}{u + 1} : u^{4} - u^{3} + u^{2} - u + 1 - \frac{1}{u + 1}

= 2 \cdot \int u^{4} - u^{3} + u^{2} - u + 1 - \frac{1}{u + 1} 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^{4} d u - \int u^{3} d u + \int u^{2} d u - \int u d u + \int 1 d u - \int \frac{1}{u + 1} d u)

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

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

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

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

\int 1 d u = u

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= 2 (\frac{u ^{5}}{5} - \frac{u ^{4}}{4} + \frac{u ^{3}}{3} - \frac{u ^{2}}{2} + u - ln ∣ u + 1 ∣)

Substitute back

u = x

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

Simplify

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

= \frac{2}{5} x^{\frac{5}{2}} - \frac{x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} - x + 2 x - 2 ln x + 1

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} - \frac{x ^{2}}{2} + \frac{2}{3} x^{\frac{3}{2}} - x + 2 x - 2 ln x + 1 + C

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