What is the integral of t^7e^{t^4} ?

The integral of t^7e^{t^4} is 1/4 (e^{t^4}t^4-e^{t^4})+C

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integral of t^7e^{t^4}

\int t^{7} e^{t^{4}} d t

\frac{1}{4} (e^{t^{4}} t^{4} - e^{t^{4}}) + C

Solution steps

\int t^{7} e^{t^{4}} d t

Apply u-substitution

: \int \frac{e ^{u^{\frac{4}{7}}} u ^{\frac{1}{7}}}{7} d u

= \int \frac{e ^{u^{\frac{4}{7}}} u ^{\frac{1}{7}}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int e^{u^{\frac{4}{7}}} u^{\frac{1}{7}} d u

Apply u-substitution

: \int \frac{7 e ^{v} v}{4} d v

= \frac{1}{7} \cdot \int \frac{7 e ^{v} v}{4} d v

Take the constant out:

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int e^{v} v d v

Apply Integration By Parts

: e^{v} v - \int e^{v} d v

= \frac{1}{7} \cdot \frac{7}{4} (e^{v} v - \int e^{v} d v)

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

= \frac{1}{7} \cdot \frac{7}{4} (e^{v} v - e^{v})

Substitute back

= \frac{1}{7} \cdot \frac{7}{4} (e^{(t^{7})^{\frac{4}{7}}} (t^{7})^{\frac{4}{7}} - e^{(t^{7})^{\frac{4}{7}}})

Simplify

\frac{1}{7} \cdot \frac{7}{4} (e^{(t^{7})^{\frac{4}{7}}} (t^{7})^{\frac{4}{7}} - e^{(t^{7})^{\frac{4}{7}}}) : \frac{1}{4} (e^{t^{4}} t^{4} - e^{t^{4}})

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

Add a constant to the solution

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

\int x^{2} \cdot e^{- 2 x} d x

\int \frac{3 x ^{2} - 5 x}{x} d x

x \to \infty lim (2 + \frac{1}{x})

y = x sin (2 x)

\int \frac{12}{1 - cos ( 2 x )} d x

What is the integral of t^7e^{t^4} ?
The integral of t^7e^{t^4} is 1/4 (e^{t^4}t^4-e^{t^4})+C