What is the integral of t^4e^{t^5+1} ?

The integral of t^4e^{t^5+1} is 1/5 e^{1+t^5}+C

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integral of t^4e^{t^5+1}

\int t^{4} e^{t^{5} + 1} d t

\frac{1}{5} e^{1 + t^{5}} + C

Solution steps

\int t^{4} e^{t^{5} + 1} d t

t^{4} e^{t^{5} + 1} = e^{1} t^{4} e^{t^{5}}

= \int e^{1} t^{4} e^{t^{5}} d t

Take the constant out:

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

= e^{1} \cdot \int t^{4} e^{t^{5}} d t

Simplify

= e \cdot \int t^{4} e^{t^{5}} d t

Apply u-substitution

: \int \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

= e \cdot \int \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= e \frac{1}{4} \cdot \int e^{u^{\frac{5}{4}}} u^{\frac{1}{4}} d u

Apply u-substitution

: \int \frac{4 e ^{v}}{5} d v

= e \frac{1}{4} \cdot \int \frac{4 e ^{v}}{5} d v

Take the constant out:

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

= e \frac{1}{4} \cdot \frac{4}{5} \cdot \int e^{v} d v

Use the common integral:

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

= e \frac{1}{4} \cdot \frac{4}{5} e^{v}

Substitute back

= e \frac{1}{4} \cdot \frac{4}{5} e^{(t^{4})^{\frac{5}{4}}}

Simplify

e \frac{1}{4} \cdot \frac{4}{5} e^{(t^{4})^{\frac{5}{4}}} : \frac{1}{5} e^{1 + t^{5}}

= \frac{1}{5} e^{1 + t^{5}}

Add a constant to the solution

= \frac{1}{5} e^{1 + t^{5}} + C

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What is the integral of t^4e^{t^5+1} ?
The integral of t^4e^{t^5+1} is 1/5 e^{1+t^5}+C