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

The integral of sqrt(4+4x^2+x^4) is x(x^2+2)-(2x^3)/3+C

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

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

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

Solution steps

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

Simplify

x^{2} : xx

= \int 4 + 4 xx + x^{4} d x

Apply Integration By Parts

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

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

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

= x 4 + 4 x^{2} + x^{4} - \frac{2 x ^{3}}{3}

4 + 4 x^{2} + x^{4} = x^{2} + 2

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

Add a constant to the solution

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

\frac{6}{( s ^{2} \cdot ( s ^{2} + 8 s + 1 ) )}

\frac{d}{d x} (e^{2 x} - 16)

\frac{d}{d x} (5 x + 6)

\int 7 csc^{2} (x) d x

\int r^{2} (r^{3} + 6)^{8} d r

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