Integration of trigonometric function

[Agent M27]

Homework Statement

$$\int$$$$\frac{2x}{x^{2}+6x+13}$$dx

Homework Equations

$$\int$$$$\frac{1}{u^{2}+a^{2}}$$du = $$\frac{1}{a}$$arctan($$\frac{u}{a}$$+c

$$\int$$$$\frac{du}{u}$$=ln(u) + c

The Attempt at a Solution

u=x+3
2xdu=2x dx
x=u-3

$$\int$$$$\frac{2x}{u^{2}+a^{2}}$$du
=$$\int$$$$\frac{2u-6}{u^{2}+a^{2}}$$du
=$$\int$$$$\frac{2u}{u^{2}+a^{2}}$$du + $$\int$$$$\frac{-6}{u^{2}+a^{2}}$$du

This is where I get stuck because I can only figure out the second integral which turns out to be:
-3arctan($$\frac{x+3}{2}$$+c

Basically for the first integral I want to know if this is allowed?

v=u$$^{2}$$+a$$^{2}$$
dv=2u du

$$\int$$$$\frac{dv}{v}$$=ln(v)

ln(v)=ln(u$$^{2}+a^{2}$$)=ln((x+3)$$^{2}$$+4)+c

So my answer would be the following, which agrees with the book:

ln(x$$^{2}$$+6x+13)-3arctan($$\frac{u}{a}$$+c

 

[Subdot]

Basically for the first integral I want to know if this is allowed?

Without checking over your work (which I don't have time to do), I can say that yes, that is allowed. You can always substitute as much as you want so long as you give the answer in the desired form in the end.

 

[rbnphlp]

Homework Statement

$$\int$$$$\frac{2x}{x^{2}+6x+13}$$dx

Homework Equations

$$\int$$$$\frac{1}{u^{2}+a^{2}}$$du = $$\frac{1}{a}$$arctan($$\frac{u}{a}$$+c

$$\int$$$$\frac{du}{u}$$=ln(u) + c

The Attempt at a Solution

u=x+3
2xdu=2x dx
x=u-3

$$\int$$$$\frac{2x}{u^{2}+a^{2}}$$du
=$$\int$$$$\frac{2u-6}{u^{2}+a^{2}}$$du
=$$\int$$$$\frac{2u}{u^{2}+a^{2}}$$du + $$\int$$$$\frac{-6}{u^{2}+a^{2}}$$du

This is where I get stuck because I can only figure out the second integral which turns out to be:
-3arctan($$\frac{x+3}{2}$$+c

Basically for the first integral I want to know if this is allowed?

v=u$$^{2}$$+a$$^{2}$$
dv=2u du

$$\int$$$$\frac{dv}{v}$$=ln(v)

ln(v)=ln(u$$^{2}+a^{2}$$)=ln((x+3)$$^{2}$$+4)+c

So my answer would be the following, which agrees with the book:

ln(x$$^{2}$$+6x+13)-3arctan($$\frac{u}{a}$$+c

just remeber
$$\int$$$$\frac{f'(x)}{f(x)}dx=lnf(x)$$

 

[Mark44]

just remeber
$$\int\frac{f'(x)}{f(x)}dx=lnf(x)$$

Don't forget the absolute value and the constant of integration.
$$\int\frac{f'(x)}{f(x)}dx=ln|f(x)| + C$$

 

[rbnphlp]

mehh ,whatever