karush said:
$\tiny{206.07.05.88}$
\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
$\displaystyle \begin{align*} \int{ \frac{1}{\left( \sqrt{u} \right) ^2 + 1} \left( \frac{1}{2\,\sqrt{u}} \right) \,\mathrm{d}u } &= \int{ \frac{1}{\tan^2{ \left( \theta \right) } + 1}\,\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= \int{ \frac{\sec^2{ \left( \theta \right) } }{\sec^2{ \left( \theta \right) }} \,\mathrm{d}\theta } \\ &= \int{
1\,\mathrm{d}\theta } \\ &= \theta + C \\ &= \arctan{ \left( \sqrt{u} \right) } + C \\ &= \arctan{ \left( \sqrt{ x^2 + 4\,x + 3 } \right) } + C \end{align*}$