What is the integral of t^3e^{t^4} ?

The integral of t^3e^{t^4} is 1/4 e^{t^4}+C

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integral of t^3e^{t^4}

\int t^{3} e^{t^{4}} d t

\frac{1}{4} e^{t^{4}} + C

Solution steps

\int t^{3} e^{t^{4}} d t

Apply u-substitution

: \int \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u^{\frac{4}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int \frac{3 e ^{v}}{4} d v

= \frac{1}{3} \cdot \int \frac{3 e ^{v}}{4} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{4} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{3} \cdot \frac{3}{4} e^{v}

Substitute back

= \frac{1}{3} \cdot \frac{3}{4} e^{(t^{3})^{\frac{4}{3}}}

Simplify

\frac{1}{3} \cdot \frac{3}{4} e^{(t^{3})^{\frac{4}{3}}} : \frac{1}{4} e^{t^{4}}

= \frac{1}{4} e^{t^{4}}

Add a constant to the solution

= \frac{1}{4} e^{t^{4}} + C

y^{^{''}} + 16 y = 2 sec (4 t)

f (x) = \frac{9}{x}

\int (sin^{2} (2 x) cos^{4} (2 x)) d x

x \to - 6 lim (f (x))

\frac{d}{d x} (cos^{2} (4 x + 1))

What is the integral of t^3e^{t^4} ?
The integral of t^3e^{t^4} is 1/4 e^{t^4}+C