What is the integral of (x+5)sqrt(10x+x^2) ?

The integral of (x+5)sqrt(10x+x^2) is 1/3 (10x+x^2)^{3/2}+C

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integral of (x+5)sqrt(10x+x^2)

\int (x + 5) 10 x + x^{2} d x

\frac{1}{3} (10 x + x^{2})^{\frac{3}{2}} + C

Solution steps

\int (x + 5) 10 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 = 10 x + x^{2}

= \frac{( 10 x + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 10 x + x ^{2} ) ^{3}}{3} : \frac{1}{3} (10 x + x^{2})^{\frac{3}{2}}

= \frac{1}{3} (10 x + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (10 x + x^{2})^{\frac{3}{2}} + C

f (x) = x + \frac{4}{x}, (4, 5)

\frac{\partial}{\partial x} (y sin (x y))

\frac{d}{d x} (- \frac{36}{( 1 + 6 x ) ^{2}})

\frac{d y}{d x} = \frac{9 x ^{2} - 6}{x ^{2}}

\frac{\partial}{\partial y} (x e^{y} - x^{5} + y - 3)

What is the integral of (x+5)sqrt(10x+x^2) ?
The integral of (x+5)sqrt(10x+x^2) is 1/3 (10x+x^2)^{3/2}+C