Verified answer

ANSWER:

[tex] \boxed{\boldsymbol{\rm{\dfrac{2}{\sqrt{3}}\arctan\left(\dfrac{x}{\sqrt{3x + 3}}\right) + C}}} [/tex]

PROBLEM:

Evaluate the indefinite integral

[tex] \displaystyle \rm \int \dfrac{x + 2}{(x^2 + 3x + 3)\sqrt{x + 1}}\ dx [/tex]

SOLUTION:

We can rewrite the given integral into

[tex] \displaystyle \rm \int \dfrac{ x + 2 }{((x + 1)(x + 2) + 1) \sqrt{x + 1}}\ dx [/tex]

By substitution, let [tex] \rm u = \sqrt{x + 1}.[/tex]

[tex] \begin{aligned} \rm u^2 &= \rm x + 1 \implies 2u\ du = dx \\ \rm u^2 + 1 &= \rm x + 2 \end{aligned} [/tex]

The integral becomes

[tex] \displaystyle \rm \int \dfrac{u^2 + 1}{(u^2(u^2 + 1)+1)\cancel{u}}\cdot 2\cancel{u}\ du [/tex]

[tex] = \displaystyle \rm 2 \int \dfrac{u^2 + 1}{u^4 + u^2 + 1}\ du [/tex]

Now, divide both numerator and denominator by [tex] \rm u^2. [/tex]

[tex] = \displaystyle \rm 2 \int \dfrac{1 + \dfrac{1}{u^2}}{u^2 + 1 + \dfrac{1}{u^2}}\ du [/tex]

[tex] = \displaystyle \rm 2 \int \dfrac{1 + \dfrac{1}{u^2}}{\left(u - \dfrac{1}{u}\right)^2 + 3}\ du [/tex]

[tex] \footnotesize \because \begin{array}{l} \rm u^2 + 1 + \dfrac{1}{u^2} = \left(u - \dfrac{1}{u}\right)^2 + 3 \quad and; \\ \\ \rm \dfrac{d}{du}\left(u - \dfrac{1}{u}\right) = 1 + \dfrac{1}{u^2} \end{array} [/tex]

[tex] = \rm \dfrac{2}{\sqrt{3}}\arctan{\left(\dfrac{u - \dfrac{1}{u}}{\sqrt{3}}\right)} + C [/tex]

[tex] = \rm \dfrac{2}{\sqrt{3}}\arctan{\left(\dfrac{\sqrt{x + 1} - \dfrac{1}{\sqrt{x + 1}}}{\sqrt{3}}\right)} + C [/tex]

Simplifying, we get

[tex] = \rm \dfrac{2}{\sqrt{3}}\arctan{\left(\dfrac{x}{\sqrt{3x + 3}}\right)} + C [/tex]

Thus,

[tex] \footnotesize \boxed{\displaystyle \rm{\int \dfrac{x + 2}{(x^2 + 3x + 3)\sqrt{x + 1}}\ dx = \dfrac{2}{\sqrt{3}}\arctan{\left(\dfrac{x}{\sqrt{3x + 3}}\right)} + C}} [/tex]