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

The integral of sqrt(x^2+4x+4) is ((x+2)^2)/2+C

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

\int x^{2} + 4 x + 4 d x

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

Solution steps

\int x^{2} + 4 x + 4 d x

Complete the square

x^{2} + 4 x + 4 : (x + 2)^{2}

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

u = x + 2

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

Add a constant to the solution

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

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