How do I go about integrating this?

[JFonseka]

Homework Statement

Find $$\int$$$$\frac{x+2}{x^{2}+2*x+10}$$*dx

You are given that $$\int$$$$\frac{du}{u^{2}+a^{2}}$$ = $$\frac{1}{a}$$tan$$^{-1}$$ $$\frac{u}{a}$$ + C for a not equal to 0

Homework Equations

None.

The Attempt at a Solution

I'm not entirely too sure how to go about doing this. I first thought of integration by parts, but given that this is a 3 mark question it seems quite long winded.

I split up the expression:

1/x^2+2*x+10*x+2*dx

Then I set u = x+2, du = 1, dv =1/x^2+2*x+10, v = (1/3)*arctan((1/3)*x+1/3)

I have no idea how to get v, that was calculated using Maple, so how is that arctan component calculated by hand?

Using the uv - int(vdu) formula for the rest of it results in some bizarre answers, which might be right but seems altogether too long and complex for a 3 mark answer.The final answer as calculated by Maple is:

(1/2)*ln(x^2+2*x+10)+(1/3)*arctan((1/3)*x+1/3)

Any help and direction greatly appreciated.

 

[HallsofIvy]

Homework Statement

Find $$\int$$$$\frac{x+2}{x^{2}+2*x+10}$$*dx

You are given that $$\int$$$$\frac{du}{u^{2}+a^{2}}$$ = $$\frac{1}{a}$$tan$$^{-1}$$ $$\frac{u}{a}$$ + C for a not equal to 0

Homework Equations

None.

The Attempt at a Solution

I'm not entirely too sure how to go about doing this. I first thought of integration by parts, but given that this is a 3 mark question it seems quite long winded.

I split up the expression:

1/x^2+2*x+10*x+2*dx

Please use parentheses to show what you mean! I finally realized that this was (1/(x^2+ 2x+ 10))(x+2). What you wrote would be more readily interpreted as 1/(x^2+ 2x+ 10x+ 2).

Then I set u = x+2, du = 1, dv =1/x^2+2*x+10, v = (1/3)*arctan((1/3)*x+1/3)

I have no idea how to get v, that was calculated using Maple, so how is that arctan component calculated by hand?

Don't do this "by parts", complete the square in the denominator. $x^2+ 2x+ 10= x^2+ 2x+ 1- 1+10= (x+1)^2+ 9$. Now you can let u= x+1 so that du= dx and x= u- 1. The integral becomes
$$\int \frac{u- 1}{u^2+ 9} du= \int \frac{u}{u^2+9} du- \int \frac{1}{u^2+9}du$$
The first integral can be done with the substitution $v= u^2+9$ and 1/(u^2+9) is exactly what you need for your arctan integral.

Using the uv - int(vdu) formula for the rest of it results in some bizarre answers, which might be right but seems altogether too long and complex for a 3 mark answer.


The final answer as calculated by Maple is:

(1/2)*ln(x^2+2*x+10)+(1/3)*arctan((1/3)*x+1/3)

Any help and direction greatly appreciated.

 

[JFonseka]

HallsofIvy, thank you so much for your response, that was a lot of help!

Sorry about the lack of parentheses.