What is the integral of x^2sqrt(x^2+3) ?

The integral of x^2sqrt(x^2+3) is

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integral of x^2sqrt(x^2+3)

\int x^{2} x^{2} + 3 d x

9 (- \frac{1}{4} (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + \frac{1}{36} x (x^{2} + 3)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} x^{2} + 3 d x

Apply Trigonometric Substitution

= \int 9 sec^{3} (u) tan^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

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

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 9 \cdot \int - sec^{3} (u) + sec^{5} (u) 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

= 9 (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

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

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= 9 (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = arctan (\frac{1}{3} x)

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

Simplify

9 - (\frac{1}{2} sec (arctan (\frac{1}{3} x)) tan (arctan (\frac{1}{3} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{3} x ) ) sin ( arctan ( \frac{1}{3} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{3} x)) tan (arctan (\frac{1}{3} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))) : 9 (- \frac{1}{4} (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + \frac{1}{36} x (x^{2} + 3)^{\frac{3}{2}})

= 9 (- \frac{1}{4} (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + \frac{1}{36} x (x^{2} + 3)^{\frac{3}{2}})

Add a constant to the solution

= 9 (- \frac{1}{4} (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + \frac{1}{36} x (x^{2} + 3)^{\frac{3}{2}}) + C

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