How do I evaluate $\int \frac {x+4}{ 2x+6 } dx $?

Question

$$\int \frac {x+4}{ 2x+6 } dx$$

This is a problem from Khan Academy that I was reading about how to solve when I accidentally clicked next and lost the explanation. I was reading something about how there is a clever way to divide the function to make it easier to integrate. Can someone please explain this to me?

No actual solution, please. I want to get it by myself.

Answer 1

HINT: Notice $$\int\frac{x+4}{2x+6}dx$$ $$=\int\frac{x+4}{2(x+3)}dx$$ $$=\frac{1}{2}\int\frac{x+4}{x+3}dx$$ $$=\frac{1}{2}\int\frac{(x+3)+1}{x+3}dx$$

$$=\frac{1}{2}\int\left(1+\frac{1}{x+3}\right)dx$$

Answer 2

The way you have to think about it is like this. $$\int \frac{x+4}{2x+6}\ dx$$ then factor the denominator to get $2(x+3)$ then pull one half out the integral to get $$ \frac{1}{2}\int \frac{x+4}{x+3}\ dx$$. here is the cleverness rewrite $x+4$ as $x+3+1$ then simplify to get $$ \frac{1}{2}\int \left(1+ \frac{1}{x+3}\right) dx$$ Hope that helps.

Answer 3

By chain rule and fundamental theorem of calculus we have $$ \int_{x} \frac{x+4}{2x+6} = \frac{1}{2} \int_{x} \frac{2x+8}{2x+6} = \frac{1}{2} \int_{x} \left( 1 + \frac{1}{x+3} \right) = \frac{x}{2} + \frac{1}{4}\int_{x}\frac{D(x+3)}{x+3} + \text{some constant} \\= \frac{x}{2} + \frac{1}{2}\ln |x+3| + \text{some constant}. $$

Answer 4

After seeing $ 2 x+ 6 $ in denominator and then the moment you see $x+4$ in numerator it should be doubled and halved as..

$$\int\frac{x+4}{2x+6}dx$$

$$=\frac12 \int\frac{2 x+8}{(2 x + 6 )}dx$$

$$=\frac{x}{2}+ \int\frac{dx}{2 x + 6}$$

$$=\frac{x}{2}+ \frac12 \int\frac{dx}{ x + 3},$$

it can be continued.

Answer 5

Singular behavior at $x = -3$:

$$\frac{x+4}{2x+6} = \frac{1}{2}\frac{1}{x+3} +\mathcal{O}(1)$$

Singular behavior at infinity:

$$\frac{x+4}{2x+6} = \frac{1}{2} + \mathcal{O}\left(\frac{1}{x}\right)$$

The sum of all the singular parts of all the expansions around all singularities equals the rational function:

$$\frac{x+4}{2x+6} = \frac{1}{2} + \frac{1}{2}\frac{1}{x+3}$$

This is true because the difference between the sum of all the singular parts of all the expansions around all singularities and the rational function has no singularities left including at infinity, which means that it is (up to removable singularities) a polynomial that tends to zero at infinity. This means that it is actually equal to zero everywhere.