What is the integral of (x+18)sqrt(36x+x^2) ?

The integral of (x+18)sqrt(36x+x^2) is 1/3 (36x+x^2)^{3/2}+C

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integral of (x+18)sqrt(36x+x^2)

\int (x + 18) 36 x + x^{2} d x

\frac{1}{3} (36 x + x^{2})^{\frac{3}{2}} + C

Solution steps

\int (x + 18) 36 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 = 36 x + x^{2}

= \frac{( 36 x + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 36 x + x ^{2} ) ^{3}}{3} : \frac{1}{3} (36 x + x^{2})^{\frac{3}{2}}

= \frac{1}{3} (36 x + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (36 x + x^{2})^{\frac{3}{2}} + C

f (x) = (x - 2)^{2}

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

\frac{\partial}{\partial x} (e^{- 2 y} sin (2 x))

\int \frac{9 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

\frac{d}{d x} ((\frac{4 x ^{2}}{2 - x})^{3})

What is the integral of (x+18)sqrt(36x+x^2) ?
The integral of (x+18)sqrt(36x+x^2) is 1/3 (36x+x^2)^{3/2}+C