Evaluate the integral: int a^{3x + 3} dx. | vedclass

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(B) To evaluate the integral $I = \int a^{3x + 3} dx$,we use the method of substitution.Let $t = 3x + 3$.Then,differentiating both sides with respect

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$\frac{a^{3x + 3}}{3 \log a} + c$

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(B) To evaluate the integral $I = \int a^{3x + 3} dx$,we use the method of substitution.Let $t = 3x + 3$.Then,differentiating both sides with respect to $x$,we get $dt = 3 dx$,which implies $dx = \frac{1}{3} dt$.Substituting these into the integral:$I = \int a^t \cdot \frac{1}{3} dt = \frac{1}{3} \int a^t dt$.Using the standard integration formula $\int a^t dt = \frac{a^t}{\log_e a} + c$,we get:$I = \frac{1}{3} \cdot \frac{a^t}{\log_e a} + c$.Finally,substituting $t = 3x + 3$ back into the expression:$I = \frac{a^{3x + 3}}{3 \log_e a} + c$.

Evaluate the integral: $\int a^{3x + 3} dx$.

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