What is the integral of (t+4)e^{t^2+8t+1} ?

The integral of (t+4)e^{t^2+8t+1} is 1/2 e^{1+t^2+8t}+C

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integral of (t+4)e^{t^2+8t+1}

\int (t + 4) e^{t^{2} + 8 t + 1} d t

\frac{1}{2} e^{1 + t^{2} + 8 t} + C

Solution steps

\int (t + 4) e^{t^{2} + 8 t + 1} d t

(t + 4) e^{t^{2} + 8 t + 1} = e^{1} (t + 4) e^{t^{2} + 8 t}

= \int e^{1} (t + 4) e^{t^{2} + 8 t} 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^{2} + 8 t} d t

Simplify

= e \cdot \int (t + 4) e^{t^{2} + 8 t} d t

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= e \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= e \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= e \frac{1}{2} e^{u}

Substitute back

u = t^{2} + 8 t

= e \frac{1}{2} e^{t^{2} + 8 t}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{t^{2} + 8 t} = e^{1 + t^{2} + 8 t}

= \frac{1}{2} e^{1 + t^{2} + 8 t}

Add a constant to the solution

= \frac{1}{2} e^{1 + t^{2} + 8 t} + C

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What is the integral of (t+4)e^{t^2+8t+1} ?
The integral of (t+4)e^{t^2+8t+1} is 1/2 e^{1+t^2+8t}+C