What is the integral of x^5sqrt(1+x^2) ?

The integral of x^5sqrt(1+x^2) is 1/7 (1+x^2)^{7/2}-2/5 (1+x^2)^{5/2}+1/3 (1+x^2)^{3/2}+C

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integral of x^5sqrt(1+x^2)

\int x^{5} 1 + x^{2} d x

\frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{5} 1 + x^{2} d x

Apply Trigonometric Substitution

: \int tan^{5} (u) sec^{3} (u) d u

= \int tan^{5} (u) sec^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} (v^{2} - 1)^{2} d v

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

Expand

v^{2} (v^{2} - 1)^{2} : v^{6} - 2 v^{4} + v^{2}

= \int v^{6} - 2 v^{4} + v^{2} 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

= \int v^{6} d v - \int 2 v^{4} d v + \int v^{2} d v

\int v^{6} d v = \frac{v ^{7}}{7}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

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

= \frac{v ^{7}}{7} - \frac{2 v ^{5}}{5} + \frac{v ^{3}}{3}

Substitute back

= \frac{sec ^{7} ( arctan ( x ) )}{7} - \frac{2 sec ^{5} ( arctan ( x ) )}{5} + \frac{sec ^{3} ( arctan ( x ) )}{3}

Simplify

\frac{sec ^{7} ( arctan ( x ) )}{7} - \frac{2 sec ^{5} ( arctan ( x ) )}{5} + \frac{sec ^{3} ( arctan ( x ) )}{3} : \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}}

= \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{7} (1 + x^{2})^{\frac{7}{2}} - \frac{2}{5} (1 + x^{2})^{\frac{5}{2}} + \frac{1}{3} (1 + x^{2})^{\frac{3}{2}} + C

y^{^{''}} = k y

\int \frac{x ^{3}}{x ^{2} + 16} d x

n = 1 \sum \infty \frac{( x ^{n} )}{n !}

sin^{2} (1 - s^{2})

What is the integral of x^5sqrt(1+x^2) ?
The integral of x^5sqrt(1+x^2) is 1/7 (1+x^2)^{7/2}-2/5 (1+x^2)^{5/2}+1/3 (1+x^2)^{3/2}+C