Recall from the Trig Substitution section that in order to do a trig substitution here we first needed to complete the square on the quadratic. This gives,

$${x^2} + 4x + 5 = {x^2} + 4x + 4 - 4 + 5 = {\left( {x + 2} \right)^2} + 1$$

After completing the square the integral becomes,

$$\int{{\sqrt {{x^2} + 4x + 5} \,dx}} = \int{{\sqrt {{{\left( {x + 2} \right)}^2} + 1} \,dx}}$$

Upon doing this we can identify the trig substitution that we need. Here it is,

$$x + 2 = \tan \theta \hspace{0.5in}x = \tan \theta - 2\hspace{0.5in}dx = {\sec ^2}\theta \,d\theta$$ $$\sqrt {{{\left( {x + 2} \right)}^2} + 1} = \sqrt {{{\tan }^2}\theta + 1} = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right| = \sec \theta$$

Recall that since we are doing an indefinite integral we can drop the absolute value bars. Using this substitution the integral becomes,

$$\begin{align*}\int{{\sqrt {{x^2} + 4x + 5} \,dx}} & = \int{{{{\sec }^3}\theta \,d\theta }}\\ & = \frac{1}{2}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) + c\end{align*}$$

We can finish the integral out with the following right triangle.

$$\tan \theta = \frac{{x + 2}}{1}\hspace{0.5in}\sec \theta = \frac{{\sqrt {{x^2} + 4x + 5} }}{1} = \sqrt {{x^2} + 4x + 5}$$ $$\int{{\sqrt {{x^2} + 4x + 5} \,dx}} = \frac{1}{2}\left( {\left( {x + 2} \right)\sqrt {{x^2} + 4x + 5} + \ln \left| {x + 2 + \sqrt {{x^2} + 4x + 5} } \right|} \right) + c$$

Okay, this doesn’t factor so partial fractions just won’t work on this. Likewise, since the numerator is just “1” we can’t use the substitution $u = 2{x^2} - 3x + 2$. So, let’s see what happens if we complete the square on the denominator.

$$\begin{align*}2{x^2} - 3x + 2 & = 2\left( {{x^2} - \frac{3}{2}x + 1} \right)\\ & = 2\left( {{x^2} - \frac{3}{2}x + \frac{9}{{16}} - \frac{9}{{16}} + 1} \right)\\ & = 2\left( {{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{7}{{16}}} \right)\end{align*}$$

With this the integral is,

$$\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} = \frac{1}{2}\int{{\frac{1}{{{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{7}{{16}}}}\,dx}}$$

Now this may not seem like all that great of a change. However, notice that we can now use the following substitution.

$$u = x - \frac{3}{4}\hspace{0.5in}du = dx$$

and the integral is now,

$$\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} = \frac{1}{2}\int{{\frac{1}{{{u^2} + \frac{7}{{16}}}}\,du}}$$

We can now see that this is an inverse tangent! So, using the formula from the start of the section we get,

$$\begin{align*}\int{{\frac{1}{{2{x^2} - 3x + 2}}\,dx}} & = \frac{1}{2}\left( {\frac{4}{{\sqrt 7 }}} \right){\tan ^{ - 1}}\left( {\frac{{4u}}{{\sqrt 7 }}} \right) + c\\ & = \frac{2}{{\sqrt 7 }}{\tan ^{ - 1}}\left( {\frac{{4x - 3}}{{\sqrt 7 }}} \right) + c\end{align*}$$

This example is a little different from the previous one. In this case we do have an $x$ in the numerator however the numerator still isn’t a multiple of the derivative of the denominator and so a simple Calculus I substitution won’t work.

So, let’s again complete the square on the denominator and see what we get,

$${x^2} + 10x + 28 = {x^2} + 10x + 25 - 25 + 28 = {\left( {x + 5} \right)^2} + 3$$

Upon completing the square the integral becomes,

$$\int{{\frac{{3x - 1}}{{{x^2} + 10x + 28}}\,dx}} = \int{{\frac{{3x - 1}}{{{{\left( {x + 5} \right)}^2} + 3}}\,dx}}$$

At this point we can use the same type of substitution that we did in the previous example. The only real difference is that we’ll need to make sure that we plug the substitution back into the numerator as well.

$$u = x + 5\hspace{0.5in}x = u - 5\hspace{0.5in}dx = du$$ $$\begin{align*}\int{{\frac{{3x - 1}}{{{x^2} + 10x + 28}}\,dx}} & = \int{{\frac{{3\left( {u - 5} \right) - 1}}{{{u^2} + 3}}\,du}}\\ & = \int{{\frac{{3u}}{{{u^2} + 3}} - \frac{{16}}{{{u^2} + 3}}\,du}}\\ & = \frac{3}{2}\ln \left| {{u^2} + 3} \right| - \frac{{16}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + c\\ & = \frac{3}{2}\ln \left| {{{\left( {x + 5} \right)}^2} + 3} \right| - \frac{{16}}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{x + 5}}{{\sqrt 3 }}} \right) + c\end{align*}$$

For the most part this integral will work the same as the previous two with one exception that will occur down the road. So, let’s start by completing the square on the quadratic in the denominator.

$${x^2} - 6x + 11 = {x^2} - 6x + 9 - 9 + 11 = {\left( {x - 3} \right)^2} + 2$$

The integral is then,

$$\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} = \int{{\frac{x}{{{{\left[ {{{\left( {x - 3} \right)}^2} + 2} \right]}^3}}}\,dx}}$$

Now, we will use the same substitution that we’ve used to this point in the previous two examples.

$$u = x - 3\hspace{0.5in}x = u + 3\hspace{0.5in}dx = du$$ $$\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = \int{{\frac{{u + 3}}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}}\\ & = \int{{\frac{u}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}} + \int{{\frac{3}{{{{\left( {{u^2} + 2} \right)}^3}}}\,du}}\end{align*}$$

Now, here is where the differences start cropping up. The first integral can be done with the substitution $v = {u^2} + 2$ and isn’t too difficult. The second integral however, can’t be done with the substitution used on the first integral and it isn’t an inverse tangent.

It turns out that a trig substitution will work nicely on the second integral and it will be the same as we did when we had square roots in the problem.

$$u = \sqrt 2 \tan \theta \hspace{0.5in}du = \sqrt 2 {\sec ^2}\theta \,d\theta$$

With these two substitutions the integrals become,

$$\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = \frac{1}{2}\int{{\frac{1}{{{v^3}}}\,dv}} + \int{{\frac{3}{{{{\left( {2{{\tan }^2}\theta + 2} \right)}^3}}}\,\left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{v^2}}} + \int{{\frac{{3\sqrt 2 {{\sec }^2}\theta }}{{8{{\left( {{{\tan }^2}\theta + 1} \right)}^3}}}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{u^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^3}}}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{\frac{1}{{{{\sec }^4}\theta }}\,d\theta }}\\ & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\int{{{{\cos }^4}\theta \,d\theta }}\end{align*}$$

Okay, at this point we’ve got two options for the remaining integral. We can either use the ideas we learned in the section about integrals involving trig integrals or we could use the following formula.

$$\int{{{{\cos }^m}\theta \,d\theta }} = \frac{1}{m}\sin \theta {\cos ^{m - 1}}\theta + \frac{{m - 1}}{m}\int{{{{\cos }^{m - 2}}\theta \,d\theta }}$$

Let’s use this formula to do the integral.

$$\begin{align*}\int{{{{\cos }^4}\theta \,d\theta }} & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{4}\int{{{{\cos }^2}\theta \,d\theta }}\\ & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{4}\left( {\frac{1}{2}\sin \theta \cos \theta + \frac{1}{2}\int{{{{\cos }^0}\theta \,d\theta }}} \right)\hspace{0.25in}{\cos ^0}\theta = 1!\\ & = \frac{1}{4}\sin \theta {\cos ^3}\theta + \frac{3}{8}\sin \theta \cos \theta + \frac{3}{8}\theta \end{align*}$$

Next, let’s use the following right triangle to get this back to $x$’s.

$$\tan \theta = \frac{u}{{\sqrt 2 }} = \frac{{x - 3}}{{\sqrt 2 }}\hspace{0.5in}\sin \theta = \frac{{x - 3}}{{\sqrt {{{\left( {x - 3} \right)}^2} + 2} }}\hspace{0.5in}\cos \theta = \frac{{\sqrt 2 }}{{\sqrt {{{\left( {x - 3} \right)}^2} + 2} }}$$

The cosine integral is then,

$$\begin{align*}\int{{{{\cos }^4}\theta \,d\theta }} & = \frac{1}{4}\frac{{2\sqrt 2 \left( {x - 3} \right)}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{3}{8}\frac{{\sqrt 2 \left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)\\ & = \frac{{\sqrt 2 }}{2}\frac{{x - 3}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)\end{align*}$$

All told then the original integral is,

$$\begin{align*}\int{{\frac{x}{{{{\left( {{x^2} - 6x + 11} \right)}^3}}}\,dx}} & = - \frac{1}{4}\frac{1}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \\ & \hspace{1.0in}\frac{{3\sqrt 2 }}{8}\left( {\frac{{\sqrt 2 }}{2}\frac{{x - 3}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{{3\sqrt 2 }}{8}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{3}{8}{{\tan }^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right)} \right)\\ & \\ & = \frac{1}{8}\frac{{3x - 11}}{{{{\left( {{{\left( {x - 3} \right)}^2} + 2} \right)}^2}}} + \frac{9}{{32}}\frac{{x - 3}}{{{{\left( {x - 3} \right)}^2} + 2}} + \frac{{9\sqrt 2 }}{{64}}{\tan ^{ - 1}}\left( {\frac{{x - 3}}{{\sqrt 2 }}} \right) + c\end{align*}$$

It’s a long and messy answer, but there it is.

As with the other problems we’ll first complete the square on the denominator.

$$4 - 2x - {x^2} = - \left( {{x^2} + 2x - 4} \right) = - \left( {{x^2} + 2x + 1 - 1 - 4} \right) = - \left( {{{\left( {x + 1} \right)}^2} - 5} \right) = 5 - {\left( {x + 1} \right)^2}$$

The integral is,

$$\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} = \int{{\frac{{x - 3}}{{{{\left[ {5 - {{\left( {x + 1} \right)}^2}} \right]}^2}}}\,dx}}$$

Now, let’s do the substitution.

$$u = x + 1\hspace{0.5in}x = u - 1\hspace{0.5in}dx = du$$

and the integral is now,

$$\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = \int{{\frac{{u - 4}}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}}\\ & = \int{{\frac{u}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}} - \int{{\frac{4}{{{{\left( {5 - {u^2}} \right)}^2}}}\,du}}\end{align*}$$

In the first integral we’ll use the substitution

$$v = 5 - {u^2}$$

and in the second integral we’ll use the following trig substitution

$$u = \sqrt 5 \sin \theta \hspace{0.5in}du = \sqrt 5 \cos \theta \,d\theta$$

Using these substitutions the integral becomes,

$$\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = - \frac{1}{2}\int{{\frac{1}{{{v^2}}}\,dv}} - \int{{\frac{4}{{{{\left( {5 - 5{{\sin }^2}\theta } \right)}^2}}}\left( {\sqrt 5 \cos \theta } \right)\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{\frac{{\cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{\frac{{\cos \theta }}{{{{\cos }^4}\theta }}\,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{4\sqrt 5 }}{{25}}\int{{{{\sec }^3}\theta \,d\theta }}\\ & = \frac{1}{2}\frac{1}{v} - \frac{{2\sqrt 5 }}{{25}}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right) + c\end{align*}$$

We’ll need the following right triangle to finish this integral out.

$$\sin \theta = \frac{u}{{\sqrt 5 }} = \frac{{x + 1}}{{\sqrt 5 }}\hspace{0.5in}\sec \theta = \frac{{\sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}\hspace{0.5in}\tan \theta = \frac{{x + 1}}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}$$

So, going back to $x$’s the integral becomes,

$$\begin{align*}\int{{\frac{{x - 3}}{{{{\left( {4 - 2x - {x^2}} \right)}^2}}}\,dx}} & = \frac{1}{2}\frac{1}{{5 - {u^2}}} - \frac{{2\sqrt 5 }}{{25}}\left( {\frac{{\sqrt 5 \left( {x + 1} \right)}}{{5 - {{\left( {x + 1} \right)}^2}}} + \ln \left| {\frac{{\sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }} + \frac{{x + 1}}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}} \right|} \right) + c\\ & = \frac{1}{{10}}\frac{{1 - 4x}}{{5 - {{\left( {x + 1} \right)}^2}}} - \frac{{2\sqrt 5 }}{{25}}\ln \left| {\frac{{x + 1 + \sqrt 5 }}{{\sqrt {5 - {{\left( {x + 1} \right)}^2}} }}} \right| + c\end{align*}$$