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

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

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

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

\frac{1}{3 2} (2 x + 1)^{3} - \frac{1}{2} (2 x + 1)^{2} + C

Solution steps

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

Factor

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

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

Apply u-substitution

: \int \frac{u ( u - 1 )}{2} - \frac{u}{2} d u

= \int \frac{u ( u - 1 )}{2} - \frac{u}{2} d u

Apply the Sum Rule:

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

= \int \frac{u ( u - 1 )}{2} d u - \int \frac{u}{2} d u

\int \frac{u ( u - 1 )}{2} d u = \frac{1}{2} (\frac{u ^{3}}{3} - \frac{u ^{2}}{2})

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

= \frac{1}{2} (\frac{u ^{3}}{3} - \frac{u ^{2}}{2}) - \frac{u ^{2}}{2 2}

Substitute back

u = 2 x + 1

= \frac{1}{2} (\frac{( 2 x + 1 ) ^{3}}{3} - \frac{( 2 x + 1 ) ^{2}}{2}) - \frac{( 2 x + 1 ) ^{2}}{2 2}

Simplify

\frac{1}{2} (\frac{( 2 x + 1 ) ^{3}}{3} - \frac{( 2 x + 1 ) ^{2}}{2}) - \frac{( 2 x + 1 ) ^{2}}{2 2} : \frac{1}{3 2} (2 x + 1)^{3} - \frac{1}{2} (2 x + 1)^{2}

= \frac{1}{3 2} (2 x + 1)^{3} - \frac{1}{2} (2 x + 1)^{2}

Add a constant to the solution

= \frac{1}{3 2} (2 x + 1)^{3} - \frac{1}{2} (2 x + 1)^{2} + C

\int \frac{cos ( θ )}{4 - sin ^{2} ( θ )} d θ

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

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