What is the integral of 2e^{2t+5} ?

The integral of 2e^{2t+5} is e^{2t+5}+C

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integral of 2e^{2t+5}

\int 2 e^{2 t + 5} d t

e^{2 t + 5} + C

Solution steps

\int 2 e^{2 t + 5} d t

Take the constant out:

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

= 2 \cdot \int e^{2 t + 5} d t

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 t + 5

= 2 \cdot \frac{1}{2} e^{2 t + 5}

Simplify

2 \cdot \frac{1}{2} e^{2 t + 5} : e^{2 t + 5}

= e^{2 t + 5}

Add a constant to the solution

= e^{2 t + 5} + C

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