What is the integral of te^{5t+pi} ?

The integral of te^{5t+pi} is 1/25 (5e^{5t+pi}t-e^{5t+pi})+C

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integral of te^{5t+pi}

\int t e^{5 t + π} d t

\frac{1}{25} (5 e^{5 t + π} t - e^{5 t + π}) + C

Solution steps

\int t e^{5 t + π} d t

Apply u-substitution

= \int \frac{ln ( u ) - π}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int ln (u) - π d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{25} (\int ln (u) d u - \int π d u)

\int ln (u) d u = u ln (u) - u

\int π d u = π u

= \frac{1}{25} (u ln (u) - u - π u)

Substitute back

u = e^{π + 5 t}

= \frac{1}{25} (e^{π + 5 t} ln (e^{π + 5 t}) - e^{π + 5 t} - π e^{π + 5 t})

Simplify

\frac{1}{25} (e^{π + 5 t} ln (e^{π + 5 t}) - e^{π + 5 t} - π e^{π + 5 t}) : \frac{1}{25} (5 e^{5 t + π} t - e^{5 t + π})

= \frac{1}{25} (5 e^{5 t + π} t - e^{5 t + π})

Add a constant to the solution

= \frac{1}{25} (5 e^{5 t + π} t - e^{5 t + π}) + C

What is the integral of te^{5t+pi} ?
The integral of te^{5t+pi} is 1/25 (5e^{5t+pi}t-e^{5t+pi})+C