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

The integral of xsqrt(x^4+2x^2-1) is

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 2 u - 1}{2} d u

= \int \frac{u ^{2} + 2 u - 1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{2} + 2 u - 1 d u

Complete the square

u^{2} + 2 u - 1 : (u + 1)^{2} - 2

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

Apply u-substitution

: \int v^{2} - 2 d v

= \frac{1}{2} \cdot \int v^{2} - 2 d v

Apply Trigonometric Substitution

: \int 2 tan^{2} (w) sec (w) d w

= \frac{1}{2} \cdot \int 2 tan^{2} (w) sec (w) d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot \int tan^{2} (w) sec (w) d w

Rewrite using trig identities

= \frac{1}{2} \cdot 2 \cdot \int (- 1 + sec^{2} (w)) sec (w) d w

Expand

(- 1 + sec^{2} (w)) sec (w) : - sec (w) + sec^{3} (w)

= \frac{1}{2} \cdot 2 \cdot \int - sec (w) + sec^{3} (w) d w

Apply the Sum Rule:

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

= \frac{1}{2} \cdot 2 (- \int sec (w) d w + \int sec^{3} (w) d w)

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

\int sec^{3} (w) d w = \frac{1}{2} sec (w) tan (w) + \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

= \frac{1}{2} \cdot 2 (- ln ∣ tan (w) + sec (w) ∣ + \frac{1}{2} sec (w) tan (w) + \frac{1}{2} ln ∣ tan (w) + sec (w) ∣)

Substitute back

= \frac{1}{2} \cdot 2 (- ln tan (arcsec (\frac{1}{2} (x^{2} + 1))) + sec (arcsec (\frac{1}{2} (x^{2} + 1))) + \frac{1}{2} sec (arcsec (\frac{1}{2} (x^{2} + 1))) tan (arcsec (\frac{1}{2} (x^{2} + 1))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} (x^{2} + 1))) + sec (arcsec (\frac{1}{2} (x^{2} + 1))))

Simplify

\frac{1}{2} \cdot 2 (- ln tan (arcsec (\frac{1}{2} (x^{2} + 1))) + sec (arcsec (\frac{1}{2} (x^{2} + 1))) + \frac{1}{2} sec (arcsec (\frac{1}{2} (x^{2} + 1))) tan (arcsec (\frac{1}{2} (x^{2} + 1))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} (x^{2} + 1))) + sec (arcsec (\frac{1}{2} (x^{2} + 1)))) : - ln \frac{1}{2} (x^{4} + 2 x^{2} - 1) + \frac{1}{2} (x^{2} + 1) + \frac{1}{4} (x^{2} + 1) x^{4} + 2 x^{2} - 1 + \frac{1}{2} ln \frac{x ^{4} + 2 x ^{2} - 1}{2} + \frac{1}{2} (x^{2} + 1)

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

Add a constant to the solution

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

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