What is the integral of (x+19)sqrt(38x+x^2) ?

The integral of (x+19)sqrt(38x+x^2) is 1/3 (38x+x^2)^{3/2}+C

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integral of (x+19)sqrt(38x+x^2)

\int (x + 19) 38 x + x^{2} d x

\frac{1}{3} (38 x + x^{2})^{\frac{3}{2}} + C

Solution steps

\int (x + 19) 38 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 = 38 x + x^{2}

= \frac{( 38 x + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 38 x + x ^{2} ) ^{3}}{3} : \frac{1}{3} (38 x + x^{2})^{\frac{3}{2}}

= \frac{1}{3} (38 x + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (38 x + x^{2})^{\frac{3}{2}} + C

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What is the integral of (x+19)sqrt(38x+x^2) ?
The integral of (x+19)sqrt(38x+x^2) is 1/3 (38x+x^2)^{3/2}+C