What is the integral of (w^5)/(sqrt(8w^2+1)) ?

The integral of (w^5)/(sqrt(8w^2+1)) is 1/960 (24w^4-4w^2+1)sqrt(8w^2+1)+C

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integral of (w^5)/(sqrt(8w^2+1))

\int \frac{w ^{5}}{8 w ^{2} + 1} d w

\frac{1}{960} (24 w^{4} - 4 w^{2} + 1) 8 w^{2} + 1 + C

Solution steps

\int \frac{w ^{5}}{8 w ^{2} + 1} d w

Apply Trigonometric Substitution

: \int \frac{tan ^{5} ( u ) sec ( u )}{512} d u

= \int \frac{tan ^{5} ( u ) sec ( u )}{512} d u

Take the constant out:

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

= \frac{1}{512} \cdot \int tan^{5} (u) sec (u) d u

Rewrite using trig identities

= \frac{1}{512} \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

Apply u-substitution

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

= \frac{1}{512} \cdot \int (- 1 + v^{2})^{2} d v

Expand

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

= \frac{1}{512} \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

= \frac{1}{512} (\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}

= \frac{1}{512} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{512} (sec (arctan (22 w)) - \frac{2 sec ^{3} ( arctan ( 2 2 w ) )}{3} + \frac{sec ^{5} ( arctan ( 2 2 w ) )}{5})

Simplify

\frac{1}{512} (sec (arctan (22 w)) - \frac{2 sec ^{3} ( arctan ( 2 2 w ) )}{3} + \frac{sec ^{5} ( arctan ( 2 2 w ) )}{5}) : \frac{1}{960} (24 w^{4} - 4 w^{2} + 1) 8 w^{2} + 1

= \frac{1}{960} (24 w^{4} - 4 w^{2} + 1) 8 w^{2} + 1

Add a constant to the solution

= \frac{1}{960} (24 w^{4} - 4 w^{2} + 1) 8 w^{2} + 1 + C

f (x) = 3 x - 2 x^{2}, (- 2, - 14)

\frac{\partial}{\partial x} (- 6 x y - x^{6} - y^{6})

\int \frac{1}{6 x + 5} d x

\int 3 x + 3 d x

(y^{4} - x^{2}) (\frac{d y}{d x}) = - x y

What is the integral of (w^5)/(sqrt(8w^2+1)) ?
The integral of (w^5)/(sqrt(8w^2+1)) is 1/960 (24w^4-4w^2+1)sqrt(8w^2+1)+C