# Thread: Integral of a rational function

1. $\textsf{evaluate}$
\begin{align}
\displaystyle
{I}&={\int{\frac{x+2}{x^2+1}dx}}\\
&=\int{\frac{x}{x^{2}{+1}}dx{\ +\ 2}\int{\frac{1}{x^{2}{+1}}}}{\ }dx\\
u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
x&=\sqrt{u-1}\\
\end{align}
...?

$\textit{calculator answer.?}$

$\dfrac{\ln\left(x^2+1\right)}{2} +2\arctan\left(x\right)+C$

2. Since $du = 2x\, dx$, then $(1/2)du = dx$. So $x\, dx/(x^2 + 1) = (1/2)du$. Take it from there.

As for the other integral, use a tangent substitution.

$\textsf{substitute$x=\tan(u)
\therefore dx=\sec^2(u)$}$
\begin{align}
\displaystyle
&=2\int\frac{\tan\left({u}\right)}
{\tan^2(u)+1}
\cdot \sec^2(u )\, du
=2\int \tan\left({u}\right) \, du
=-2\ln\left({\cos\left({u}\right)}\right)+C
\end{align}
$\textsf{substitute u=arctan{x}}$

$\textit{this doesn't seem to be headed towards the answer}$

$\textit{calculator answer}$

$\dfrac{\ln\left(x^2+1\right)}{2} +2\arctan\left(x\right)+C$

4. The tangent substitution was meant for the second integral

$$2\int \frac{1}{x^2 + 1}\, dx$$

and not the integral $\int x\, dx/(x^2 + 1)$. I already explained how to evaluate that integral with the $u$ you had already chose, namely, $u = x^2 + 1$.

so...
$\textsf{solving}\\$
\begin{align}
\displaystyle
I_a&=\int \frac{x}{x^{2}+1} \, dx \\
u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
\textit{u substitution}\\
&=\frac{1}{2} \int \frac{1}{u}du =\frac{\ln\left({u}\right)}{2}\\
\textit{backsubstition }\\
I_a&=\dfrac{\ln\left(x^2+1\right)}{2}
\end{align}
$\textsf{solving}\\$
\begin{align}
\displaystyle
I_b &= 2 \int \frac{1}{x^{2}+1} \, dx\\
x&=\tan(u) \therefore dx=\sec^2(u) du \\
u&=\arctan(x) \\
\textit{u substitution }\\
I_b &= 2 \int \frac{1}{\tan^{2}u+1} \, \sec^2(u) du \\
&=2 \int 1 du= 2u\\
\textit{backsubstition}\\
I_b&=2\arctan{(x)}
\end{align}

$\textit{$I_a + I_b +C = $}$

$\dfrac{\ln\left(x^2+1\right)}{2} +2\arctan\left(x\right)+C$

6. I would simply write:

$\displaystyle I=\int\frac{x+2}{x^2+1}\,dx=\frac{1}{2}\int \frac{2x}{x^2+1}\,dx+2\int \frac{1}{x^2+1}\,dx=\frac{1}{2}\ln(x^2+1)+2\arctan(x)+C$

7. Originally Posted by karush
so...
$\textsf{solving}\\$
\begin{align}
\displaystyle
I_a&=\int \frac{x}{x^{2}+1} \, dx \\
u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\
\textit{u substitution}\\
&=\frac{1}{2} \int \frac{1}{u}du =\frac{\ln\left({u}\right)}{2}\\
\textit{backsubstition }\\
I_a&=\dfrac{\ln\left(x^2+1\right)}{2}
\end{align}
$\textsf{solving}\\$
\begin{align}
\displaystyle
I_b &= 2 \int \frac{1}{x^{2}+1} \, dx\\
x&=\tan(u) \therefore dx=\sec^2(u) du \\
u&=\arctan(x) \\
\textit{u substitution }\\
I_b &= 2 \int \frac{1}{\tan^{2}u+1} \, \sec^2(u) du \\
&=2 \int 1 du= 2u\\
\textit{backsubstition}\\
I_b&=2\arctan{(x)}
\end{align}

$\textit{$I_a + I_b +C = $}$

$\dfrac{\ln\left(x^2+1\right)}{2} +2\arctan\left(x\right)+C$

Your work here is correct.

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