What is the integral of (x^5)/(sqrt(x^2+3)) ?

The integral of (x^5)/(sqrt(x^2+3)) is 9sqrt(3)sqrt(1/3 (3+x^2))-2(3+x^2)^{3/2}+1/5 (3+x^2)^{5/2}+C

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integral of (x^5)/(sqrt(x^2+3))

\int \frac{x ^{5}}{x ^{2} + 3} d x

93 \frac{1}{3} (3 + x^{2}) - 2 (3 + x^{2})^{\frac{3}{2}} + \frac{1}{5} (3 + x^{2})^{\frac{5}{2}} + C

Solution steps

\int \frac{x ^{5}}{x ^{2} + 3} d x

Apply Trigonometric Substitution

= \int 93 tan^{5} (u) sec (u) d u

Take the constant out:

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

= 93 \cdot \int tan^{5} (u) sec (u) d u

Rewrite using trig identities

= 93 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

Apply u-substitution

= 93 \cdot \int (- 1 + v^{2})^{2} d v

Expand

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

= 93 \cdot \int 1 - 2 v^{2} + v^{4} 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

= 93 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= 93 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 93 sec (arctan (\frac{1}{3} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{3} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{3} x ) )}{5}

Simplify

93 sec (arctan (\frac{1}{3} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{3} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{3} x ) )}{5} : 93 \frac{1}{3} (3 + x^{2}) - 2 (3 + x^{2})^{\frac{3}{2}} + \frac{1}{5} (3 + x^{2})^{\frac{5}{2}}

= 93 \frac{1}{3} (3 + x^{2}) - 2 (3 + x^{2})^{\frac{3}{2}} + \frac{1}{5} (3 + x^{2})^{\frac{5}{2}}

Add a constant to the solution

= 93 \frac{1}{3} (3 + x^{2}) - 2 (3 + x^{2})^{\frac{3}{2}} + \frac{1}{5} (3 + x^{2})^{\frac{5}{2}} + C

What is the integral of (x^5)/(sqrt(x^2+3)) ?
The integral of (x^5)/(sqrt(x^2+3)) is 9sqrt(3)sqrt(1/3 (3+x^2))-2(3+x^2)^{3/2}+1/5 (3+x^2)^{5/2}+C