How do you integrate $\int \frac{x + 5}{2 x + 3}$ $int (x+5)/(2x+3)$ using substitution?

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Answer 1 · 2016-08-23T12:02:19

$= \frac{7}{4} ln (2 x + 3) + \frac{1}{2} x + C$ $=7/4ln(2x+3) + 1/2x + C$

We can't immediately substitute into this integrand. First we have to get it into a more receptive form:

We do this with polynomial long division. It's a very simple thing to do on paper but the formatting is quite difficult on here.

$\int \frac{x + 5}{2 x + 3} d x = \int (\frac{7}{2 (2 x + 3)} + \frac{1}{2}) d x$ $int (x+5)/(2x+3)dx = int (7/(2(2x+3)) + 1/2)dx$

$= \frac{7}{2} \int \frac{d x}{2 x + 3} + \frac{1}{2} \int d x$ $=7/2int (dx)/(2x+3) + 1/2intdx$

Now for the first integral set $u = 2 x + 3 \Rightarrow d u = 2 d x$ $u = 2x+3 implies du = 2dx$

$\Rightarrow d x = \frac{d u}{2}$ $implies dx = (du)/2$

$= \frac{7}{4} \int \frac{d u}{u} + \frac{1}{2} \int d x$ $=7/4int(du)/(u) + 1/2intdx$

$= \frac{7}{4} ln (u) + \frac{1}{2} x + C$ $=7/4ln(u) + 1/2x + C$

$= \frac{7}{4} ln (2 x + 3) + \frac{1}{2} x + C$ $=7/4ln(2x+3) + 1/2x + C$

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