What is the integral of (xsqrt(x))/(x+1) ?

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

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integral of (xsqrt(x))/(x+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{4}}{u ^{2} + 1} d u

= \int \frac{2 u ^{4}}{u ^{2} + 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 ^{4}}{u ^{2} + 1} d u

Long division

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

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

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

\int 1 d u = u

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

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

Substitute back

u = x

= 2 (\frac{( x ) ^{3}}{3} - x + arctan (x))

Simplify

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

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

Add a constant to the solution

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

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