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

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

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

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

\frac{x ^{3}}{3} + x + C

Solution steps

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

Factor

x^{4} + 2 x^{2} + 1 : (x^{2} + 1)^{2}

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

(x^{2} + 1)^{2} = ((x^{2} + 1)),

assuming

(x^{2} + 1) \geq 0

= \int x^{2} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x + \int 1 d x

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

\int 1 d x = x

= \frac{x ^{3}}{3} + x

Add a constant to the solution

= \frac{x ^{3}}{3} + x + C

\frac{1}{x + 4}, 2

3 e^{\frac{1}{x}}

\int_{0}^{π} sin^{2} (8 x) d x

\int_{- 4}^{4} (3 - x) d x

x \to 0 lim ((sin (x^{2}) + 1)^{\frac{1}{5 x ^{2}}})

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