What is the integral of (x+2)sqrt(x^2+4x+1) ?

The integral of (x+2)sqrt(x^2+4x+1) is 1/3 (x^2+4x+1)^{3/2}+C

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integral of (x+2)sqrt(x^2+4x+1)

\int (x + 2) x^{2} + 4 x + 1 d x

\frac{1}{3} (x^{2} + 4 x + 1)^{\frac{3}{2}} + C

Solution steps

\int (x + 2) x^{2} + 4 x + 1 d x

Apply u-substitution

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

Apply u-substitution

= \int v^{2} d v

Apply the Power Rule

= \frac{v ^{3}}{3}

Substitute back

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

Simplify

\frac{( ( x + 2 ) ^{2} - 3 ) ^{3}}{3} : \frac{1}{3} (x^{2} + 4 x + 1)^{\frac{3}{2}}

= \frac{1}{3} (x^{2} + 4 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (x^{2} + 4 x + 1)^{\frac{3}{2}} + C

What is the integral of (x+2)sqrt(x^2+4x+1) ?
The integral of (x+2)sqrt(x^2+4x+1) is 1/3 (x^2+4x+1)^{3/2}+C