What is the integral of t^3sqrt(9t^4+1) ?

The integral of t^3sqrt(9t^4+1) is 1/54 (9t^4+1)^{3/2}+C

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integral of t^3sqrt(9t^4+1)

\int t^{3} 9 t^{4} + 1 d t

\frac{1}{54} (9 t^{4} + 1)^{\frac{3}{2}} + C

Solution steps

\int t^{3} 9 t^{4} + 1 d t

Apply u-substitution

: \int \frac{u}{36} d u

= \int \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{36} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 t^{4} + 1

= \frac{1}{36} \cdot \frac{2}{3} (9 t^{4} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{36} \cdot \frac{2}{3} (9 t^{4} + 1)^{\frac{3}{2}} : \frac{1}{54} (9 t^{4} + 1)^{\frac{3}{2}}

= \frac{1}{54} (9 t^{4} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{54} (9 t^{4} + 1)^{\frac{3}{2}} + C

\frac{d y}{d x} = 3 x^{2} (1 + y^{2})

y^{^{'}} = \frac{4 x sec ( \frac{y}{x} ) + y}{x}

y^{^{''}} + π y = 0

\int (cos^{2} (t)) d t

n = 2 \sum \infty sin (\frac{1}{n})

What is the integral of t^3sqrt(9t^4+1) ?
The integral of t^3sqrt(9t^4+1) is 1/54 (9t^4+1)^{3/2}+C