Integral of 1/3e^(3x)

\int\frac{1}{3} e^{3x}dx

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Answer

$\frac{1}{9}e^{3x}+C_0$

Step by step solution

Problem

$\int\frac{1}{3} e^{3x}dx$
1

Taking the constant out of the integral

$\frac{1}{3}\int e^{3x}dx$
2

Solve the integral $\int e^{3x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=3x \\ du=3dx\end{matrix}$
3

Isolate $dx$ in the previous equation

$\frac{du}{3}=dx$
4

Substituting $u$ and $dx$ in the integral

$\frac{1}{3}\int\frac{e^u}{3}du$
5

Taking the constant out of the integral

$\frac{1}{3}\cdot \frac{1}{3}\int e^udu$
6

Multiply $\frac{1}{3}$ times $\frac{1}{3}$

$\frac{1}{9}\int e^udu$
7

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{9}e^u$
8

Substitute $u$ back for it's value, $3x$

$\frac{1}{9}e^{3x}$
9

Add the constant of integration

$\frac{1}{9}e^{3x}+C_0$

Answer

$\frac{1}{9}e^{3x}+C_0$

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Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.21 seconds

Views:

103