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

The integral of xsqrt(x^2+2x+4) is

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

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

\frac{1}{2} x^{2} x^{2} + 2 x + 4 - \frac{1}{6} (- 3 x x^{2} + 2 x + 4 + (x^{2} + 2 x + 4)^{\frac{3}{2}} + 9 ln \frac{x ^{2} + 2 x + 4}{3} + \frac{1}{3} (x + 1) - 9 x^{2} + 2 x + 4) + C

Solution steps

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

Apply Integration By Parts

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

\int \frac{x ^{2} ( x + 1 )}{2 x ^{2} + 2 x + 4} d x = \frac{1}{6} (- 3 x x^{2} + 2 x + 4 + (x^{2} + 2 x + 4)^{\frac{3}{2}} + 9 ln \frac{x ^{2} + 2 x + 4}{3} + \frac{1}{3} (x + 1) - 9 x^{2} + 2 x + 4)

= \frac{1}{2} x^{2} x^{2} + 2 x + 4 - \frac{1}{6} (- 3 x x^{2} + 2 x + 4 + (x^{2} + 2 x + 4)^{\frac{3}{2}} + 9 ln \frac{x ^{2} + 2 x + 4}{3} + \frac{1}{3} (x + 1) - 9 x^{2} + 2 x + 4)

Add a constant to the solution

= \frac{1}{2} x^{2} x^{2} + 2 x + 4 - \frac{1}{6} (- 3 x x^{2} + 2 x + 4 + (x^{2} + 2 x + 4)^{\frac{3}{2}} + 9 ln \frac{x ^{2} + 2 x + 4}{3} + \frac{1}{3} (x + 1) - 9 x^{2} + 2 x + 4) + C