What is the integral of (e^{7t}+4t+ln(t)) ?

The integral of (e^{7t}+4t+ln(t)) is 1/7 e^{7t}+2t^2+tln(t)-t+C

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integral of (e^{7t}+4t+ln(t))

\int (e^{7 t} + 4 t + ln (t)) d t

\frac{1}{7} e^{7 t} + 2 t^{2} + t ln (t) - t + C

Solution steps

\int e^{7 t} + 4 t + ln (t) d t

Apply the Sum Rule:

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

= \int e^{7 t} d t + \int 4 t d t + \int ln (t) d t

\int e^{7 t} d t = \frac{1}{7} e^{7 t}

\int 4 t d t = 2 t^{2}

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

= \frac{1}{7} e^{7 t} + 2 t^{2} + t ln (t) - t

Add a constant to the solution

= \frac{1}{7} e^{7 t} + 2 t^{2} + t ln (t) - t + C

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What is the integral of (e^{7t}+4t+ln(t)) ?
The integral of (e^{7t}+4t+ln(t)) is 1/7 e^{7t}+2t^2+tln(t)-t+C