What is the integral of xsqrt(x^4-9) ?

The integral of xsqrt(x^4-9) is 1/4 (x^2sqrt(x^4-9)-9ln((|x^2+sqrt(x^4-9)|)/3))+C

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integral of xsqrt(x^4-9)

\int x x^{4} - 9 d x

\frac{1}{4} (x^{2} x^{4} - 9 - 9 ln (\frac{x ^{2} + x ^{4} - 9}{3})) + C

Solution steps

\int x x^{4} - 9 d x

Simplify

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

= \int x (x^{2})^{2} - 9 d x

Apply u-substitution

: \int \frac{u ^{2} - 9}{2} d u

= \int \frac{u ^{2} - 9}{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} - 9 d u

Apply Trigonometric Substitution

: \int 9 tan^{2} (v) sec (v) d v

= \frac{1}{2} \cdot \int 9 tan^{2} (v) sec (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot 9 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot 9 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= \frac{1}{2} \cdot 9 \cdot \int - sec (v) + sec^{3} (v) d v

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 9 (- \int sec (v) d v + \int sec^{3} (v) d v)

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

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

= \frac{1}{2} \cdot 9 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{1}{2} \cdot 9 (- ln tan (arcsec (\frac{1}{3} x^{2})) + sec (arcsec (\frac{1}{3} x^{2})) + \frac{1}{2} sec (arcsec (\frac{1}{3} x^{2})) tan (arcsec (\frac{1}{3} x^{2})) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x^{2})) + sec (arcsec (\frac{1}{3} x^{2})))

Simplify

\frac{1}{2} \cdot 9 (- ln tan (arcsec (\frac{1}{3} x^{2})) + sec (arcsec (\frac{1}{3} x^{2})) + \frac{1}{2} sec (arcsec (\frac{1}{3} x^{2})) tan (arcsec (\frac{1}{3} x^{2})) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x^{2})) + sec (arcsec (\frac{1}{3} x^{2}))) : \frac{1}{4} (x^{2} x^{4} - 9 - 9 ln (\frac{x ^{2} + x ^{4} - 9}{3}))

= \frac{1}{4} (x^{2} x^{4} - 9 - 9 ln (\frac{x ^{2} + x ^{4} - 9}{3}))

Add a constant to the solution

= \frac{1}{4} (x^{2} x^{4} - 9 - 9 ln (\frac{x ^{2} + x ^{4} - 9}{3})) + C

\int_{0}^{3} (x^{2} - 2 x) d x

y^{^{''}} + 11 y^{^{'}} - 28 y = 0

\int \frac{8}{x ^{5}} d x

x \to 3 lim (\frac{x ^{2} - 9}{x ^{2} - 10 x + 21})

\int \frac{1}{d x} d t

What is the integral of xsqrt(x^4-9) ?
The integral of xsqrt(x^4-9) is 1/4 (x^2sqrt(x^4-9)-9ln((|x^2+sqrt(x^4-9)|)/3))+C