206.07.05.88 partial fractions?

[karush]

$\tiny{206.07.05.88}$
\begin{align*}
\displaystyle
I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
&=?
\end{align*}

would partial fractions be best for this?

 

[MarkFL]

$\tiny{206.07.05.88}$
\begin{align*}
\displaystyle
I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
&=?
\end{align*}

would partial fractions be best for this?

I would look at a substitution like:

\(\displaystyle u=\sqrt{x^2+4x+3}\implies dx=\frac{u}{x+2}\,du\)

And the integral then becomes:

\(\displaystyle I=\int\frac{1}{u^2+1}\,du\)

 

[karush]

$\displaystyle I=\int\frac{1}{u^2+1}\,du$
$\textit{then you have the standard form}\\$
\begin{align*}\displaystyle
I_{32}&=\int\frac{1}{a^2+x^2}\,dx\\
&=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C
\end{align*}
$\textit{with $a=u$ and $x=1$ then}$
\begin{align*}\displaystyle
I&=\frac{1}{u}\tan^{-1}\left(\frac{x}{u}\right)+C
\end{align*}
$\textit{since $u=\sqrt{x^2+4x+3}$ and $x=1$}\\$
\begin{align*}\displaystyle
I&=\frac{1}{\sqrt{x^2+4x+3}}\tan^{-1}\left(\frac{1}{\sqrt{x^2+4x+3}}\right)+C
\end{align*}

Really?

 

[MarkFL]

$\displaystyle I=\int\frac{1}{u^2+1}\,du$
$\textit{then you have the standard form}\\$
\begin{align*}\displaystyle
I_{32}&=\int\frac{1}{a^2+x^2}\,dx\\
&=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C
\end{align*}
$\textit{with $a=u$ and $x=1$ then}$
\begin{align*}\displaystyle
I&=\frac{1}{u}\tan^{-1}\left(\frac{x}{u}\right)+C
\end{align*}
$\textit{since $u=\sqrt{x^2+4x+3}$}\\$
\begin{align*}\displaystyle
I&=\frac{1}{\sqrt{x^2+4x+3}}\tan^{-1}\left(\frac{x}{\sqrt{x^2+4x+3}}\right)+C
\end{align*}

Really?

In the formula you've used, we have $a=1$...and so after back-substituting for $u$:

\(\displaystyle I=\arctan\left(\sqrt{x^2+4x+3}\right)+C\)

 

[karush]

did you flip

$u^2 + 1 $ to $1+u^2 $

to do this??

 

[MarkFL]

did you flip

$u^2 + 1 $ to $1+u^2 $

to do this??

No, but either way will work, because of the commutative property of addition.. :D

 

[karush]

so that is because

$u$ is in terms of $x$

so $a$ has to be $1$ ?

 

[HallsofIvy]

It looks like you are trying to use integration formulas without understanding them.

Yes, $$\int\frac{dx}{x^2+ a^2}= arctan(x/a)+ C$$[/tex].]

From that it also follows that $$\int\frac{dy}{y^2+ a^2}= arctan(y/a)+ C$$,

$$\int\frac{dt}{t^2+ a^2}= arctan(t/a)+ C$$ and even

$$\int\frac{du}{u^2+ a^2}= arctan(u/a+ C$$.

Certainly, you should have learned that "a+ b= b+ a" back in basic arithmetic even if no one ever used the word "commutative".

 

[Prove It]

$\tiny{206.07.05.88}$
\begin{align*}
\displaystyle
I_{88}&=\int\frac{1}{(x+2)\sqrt{x^2+4x+3}} \, dx \\
&=?
\end{align*}

would partial fractions be best for this?

$\displaystyle \begin{align*} \int{ \frac{1}{ \left( x + 2 \right) \sqrt{x^2 + 4\,x + 3} }\,\mathrm{d}x} &= \int{ \frac{2\,x + 4}{\left( 2\,x + 4 \right) \left( x + 2 \right) \sqrt{ x^2 + 4\,x + 3}}\,\mathrm{d}x } \\ &= \int{ \frac{2\, x + 4 }{2 \left( x + 2 \right) \left( x + 2 \right) \sqrt{x^2 + 4\,x + 3} } \,\mathrm{d}x } \\ &= \frac{1}{2} \int{ \frac{2\,x + 4}{\left( x^2 + 4\,x + 4 \right) \sqrt{x^2 + 4\,x + 3}} \,\mathrm{d}x } \\ &= \frac{1}{2} \int{ \frac{2\,x + 4}{\left( x^2 + 4\,x + 3 + 1 \right) \sqrt{x^2 + 4\,x + 3}} \,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u= x^2 + 4\,x + 3 \implies \mathrm{d}u = \left( 2\,x + 4 \right) \, \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2} \int{ \frac{2\,x + 4}{\left( x^2 + 4\,x + 3 + 1 \right) \sqrt{x^2 + 4\,x + 3}}\,\mathrm{d}x } &= \frac{1}{2} \int{ \frac{1}{\left( u + 1 \right) \sqrt{u} } \,\mathrm{d}u } \\ &= \int{ \frac{1}{ \left( \sqrt{u } \right) ^2 + 1 }\left( \frac{1}{2\,\sqrt{u}} \right) \,\mathrm{d}u } \end{align*}$

Let $\displaystyle \begin{align*} \sqrt{u } = \tan{ \left( \theta \right) } \implies \frac{1}{2\,\sqrt{u}}\,\mathrm{d}u = \sec^2{ \left( \theta \right) }\,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{1}{\left( \sqrt{u} \right) ^2 + 1} \left( \frac{1}{2\,\sqrt{u}} \right) \,\mathrm{d}u } &= \int{ \frac{1}{\tan^2{ \left( \theta \right) } + 1}\,\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= \int{ \frac{\sec^2{ \left( \theta \right) } }{\sec^2{ \left( \theta \right) }} \,\mathrm{d}\theta } \\ &= \int{
1\,\mathrm{d}\theta } \\ &= \theta + C \\ &= \arctan{ \left( \sqrt{u} \right) } + C \\ &= \arctan{ \left( \sqrt{ x^2 + 4\,x + 3 } \right) } + C \end{align*}$