What is the integral of xsqrt(x^2+6x+9) ?

The integral of xsqrt(x^2+6x+9) is ((x+3)^3)/3-(3(x+3)^2)/2+C

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integral of xsqrt(x^2+6x+9)

\int x x^{2} + 6 x + 9 d x

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

Solution steps

\int x x^{2} + 6 x + 9 d x

Complete the square

x^{2} + 6 x + 9 : (x + 3)^{2}

= \int x (x + 3)^{2} d x

Apply u-substitution

: \int (u - 3) u d u

= \int (u - 3) u d u

Expand

(u - 3) u : u^{2} - 3 u

= \int u^{2} - 3 u 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

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

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

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

= \frac{u ^{3}}{3} - \frac{3 u ^{2}}{2}

Substitute back

u = x + 3

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

Add a constant to the solution

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

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What is the integral of xsqrt(x^2+6x+9) ?
The integral of xsqrt(x^2+6x+9) is ((x+3)^3)/3-(3(x+3)^2)/2+C