What is the integral of we^{5w+pi} ?

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

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integral of we^{5w+pi}

\int w e^{5 w + π} d w

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

Solution steps

\int w e^{5 w + π} d w

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 w}

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

Simplify

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

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

Add a constant to the solution

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

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