Integrate te^(5t+pi)

\int t e^{\left(5t+\pi \right)}dt

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Answer

$-\frac{1}{25}e^{\left(\pi+5t\right)}+\frac{1}{5}te^{\left(\pi+5t\right)}+C_0$

Step by step solution

Problem

$\int t e^{\left(5t+\pi \right)}dt$
1

Use the integration by parts theorem to calculate the integral $\int te^{\left(\pi+5t\right)}dt$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=t}\\ \displaystyle{du=dt}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{\left(\pi+5t\right)}dt}\\ \displaystyle{\int dv=\int e^{\left(\pi+5t\right)}dt}\end{matrix}$
4

Solve the integral

$v=\int e^{\left(\pi+5t\right)}dt$
5

Solve the integral $\int e^{\left(\pi+5t\right)}dt$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=\pi+5t \\ du=5dt\end{matrix}$
6

Isolate $dt$ in the previous equation

$\frac{du}{5}=dt$
7

Substituting $u$ and $dt$ in the integral

$\int\frac{e^u}{5}du$
8

Taking the constant out of the integral

$\frac{1}{5}\int e^udu$
9

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 $a \neq 1$

$\frac{1}{5}e^u$
10

Substitute $u$ back for it's value, $\pi+5t$

$\frac{1}{5}e^{\left(\pi+5t\right)}$
11

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{5}\int e^{\left(\pi+5t\right)}dt$
12

Solve the integral $\int e^{\left(\pi+5t\right)}dt$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=\pi+5t \\ du=5dt\end{matrix}$
13

Isolate $dt$ in the previous equation

$\frac{du}{5}=dt$
14

Substituting $u$ and $dt$ in the integral

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{5}\int\frac{e^u}{5}du$
15

Taking the constant out of the integral

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{5}\cdot \frac{1}{5}\int e^udu$
16

Multiply $\frac{1}{5}$ times $-\frac{1}{5}$

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{25}\int e^udu$
17

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 $a \neq 1$

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{25}e^u$
18

Substitute $u$ back for it's value, $\pi+5t$

$\frac{1}{5}te^{\left(\pi+5t\right)}-\frac{1}{25}e^{\left(\pi+5t\right)}$
19

Add the constant of integration

$-\frac{1}{25}e^{\left(\pi+5t\right)}+\frac{1}{5}te^{\left(\pi+5t\right)}+C_0$

Answer

$-\frac{1}{25}e^{\left(\pi+5t\right)}+\frac{1}{5}te^{\left(\pi+5t\right)}+C_0$

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Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.36 seconds

Views:

109