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

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

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

\int (x + 1) 2 x + x^{2} d x

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

Solution steps

\int (x + 1) 2 x + x^{2} d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = 2 x + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (- \frac{2}{5 x})

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{4} + x - 2

\frac{d}{d x} (\frac{x + 2}{ln ( x )})

f (x) = x^{2} - 7 x + 7, (6, 1)

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{3 t}

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