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

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

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

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

\frac{2 x ^{3}}{3} + x + C

Solution steps

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

Factor

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

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

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

assuming

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

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

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

\int 1 d x = x

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

Add a constant to the solution

= \frac{2 x ^{3}}{3} + x + C

\int \frac{1}{t ^{2} 196 - t ^{2}} d t

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

\int 5 sin (\frac{x}{2}) d x

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

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