What is the integral of 4t^5sqrt(2t^2+3) ?

The integral of 4t^5sqrt(2t^2+3) is 2/35 (5t^4-6t^2+6)(2t^2+3)^{3/2}+C

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integral of 4t^5sqrt(2t^2+3)

\int 4 t^{5} 2 t^{2} + 3 d t

\frac{2}{35} (5 t^{4} - 6 t^{2} + 6) (2 t^{2} + 3)^{\frac{3}{2}} + C

Solution steps

\int 4 t^{5} 2 t^{2} + 3 d t

Take the constant out:

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

= 4 \cdot \int t^{5} 2 t^{2} + 3 d t

Apply Trigonometric Substitution

= 4 \cdot \int \frac{27 3}{8} tan^{5} (u) sec^{3} (u) d u

Take the constant out:

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

= 4 \cdot \frac{27 3}{8} \cdot \int tan^{5} (u) sec^{3} (u) d u

Rewrite using trig identities

= 4 \cdot \frac{27 3}{8} \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec^{3} (u) d u

Apply u-substitution

= 4 \cdot \frac{27 3}{8} \cdot \int v^{2} (v^{2} - 1)^{2} d v

Expand

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

= 4 \cdot \frac{27 3}{8} \cdot \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

= 4 \cdot \frac{27 3}{8} (\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}

= 4 \cdot \frac{27 3}{8} (\frac{v ^{7}}{7} - \frac{2 v ^{5}}{5} + \frac{v ^{3}}{3})

Substitute back

= 4 \cdot \frac{27 3}{8} \frac{sec ^{7} ( arctan ( \frac{2}{3} t ) )}{7} - \frac{2 sec ^{5} ( arctan ( \frac{2}{3} t ) )}{5} + \frac{sec ^{3} ( arctan ( \frac{2}{3} t ) )}{3}

Simplify

4 \cdot \frac{27 3}{8} \frac{sec ^{7} ( arctan ( \frac{2}{3} t ) )}{7} - \frac{2 sec ^{5} ( arctan ( \frac{2}{3} t ) )}{5} + \frac{sec ^{3} ( arctan ( \frac{2}{3} t ) )}{3} : \frac{2}{35} (5 t^{4} - 6 t^{2} + 6) (2 t^{2} + 3)^{\frac{3}{2}}

= \frac{2}{35} (5 t^{4} - 6 t^{2} + 6) (2 t^{2} + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{35} (5 t^{4} - 6 t^{2} + 6) (2 t^{2} + 3)^{\frac{3}{2}} + C

What is the integral of 4t^5sqrt(2t^2+3) ?
The integral of 4t^5sqrt(2t^2+3) is 2/35 (5t^4-6t^2+6)(2t^2+3)^{3/2}+C