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

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

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

\int t^{3} \cdot 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

\int_{1}^{t} \frac{9}{x ^{3}} d x

\frac{d}{d x} (\frac{5 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\frac{\partial}{\partial x} (sin (4 x + 4 y))

\frac{d}{d x} ((1 - x^{2})^{5})

f (x) = csc (x^{2})

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