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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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: $\int a^{3x + 3} dx$.

Similar Questions

Find the integral of the function $\sin ^{2}(2 x+5)$.

Find the following integral: $\int \left(x^{\frac{2}{3}} + 1\right) dx$

$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

If $\int \frac{dx}{32-2x^2} = A \log(4-x) + B \log(4+x) + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

$\int \frac{dx}{\sin x + \sqrt{3} \cos x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module