Integrate $\displaystyle\int{\frac{1}{2+3x}dx}$

Integrate $$\int{\frac{1}{2+3x}dx}$$

What should I do for this integral? How do I evaluate it?

Thanks!

• It's generally courteous to minimize the amount of vertical space one's question title takes up; this makes real estate distribution on the main page's list of questions fairer. As for your question, are you familiar with substitution, and do you know an antiderivative for $\frac{1}{u}$? – anon Aug 9 '13 at 14:30

HINT:

Put $2+3x=u$ so that $3dx=du$

and we know $$\int \frac{du}u=\ln |u|+C$$ where $C$ is an arbitrary constant for indefinite integration

• Oh that's right, thanks! – user89632 Aug 9 '13 at 14:34
• @user89632, my pleasure. – lab bhattacharjee Aug 9 '13 at 14:52

Use u-substitution, as lab bhattacharjee did, but I think he's skipping something important; look:

$$\int \frac{\mathrm{d}x}{2+3x} = \int \frac{\mathrm{d}u}{3u}$$

because you assumed $u = 2+3x$ and, accordingly, $\mathrm{d}u = 3\mathrm{d}x$ (thus $\mathrm{d}x = \frac{\mathrm{d}u}{3}$, as you can see from the second integral above).

Now the function whose derivative is $\frac{1}{u}$ is, as you should know, $y = \ln u$, so, if we have to take the integral of $\frac{1}{3u}$, we can write that this integral is equal to $y = \frac{1}{3}\ln u$, because we multiplied the function $\frac{1}{u}$ only by $\frac{1}{3}$, which is a constant factor, which in his turn multiplies the integral.

Finally:

$$\int \frac{\mathrm{d}u}{3u} = \frac{1}{3} \ln u + C = \frac{1}{3} \ln (2+3x) + C$$

One can also use the ‘brute-force’ method.

The intermediate target is to convert $\mathrm{d}x$ to $\mathrm{d}(3x + 2)$.

Prior to that, we need $\mathrm{d}(3x)$ first, because $\mathrm{d}(3x) = \mathrm{d}(3x + 2)$

But, $\mathrm{d}(3x) = 3\mathrm{d}(x)$. Thus, we need the ‘$3$’ comes to help. The ‘$3$’ newly introduced in the numerator can be counter-balanced by another ‘$3$’ in the denominator. In short, \begin{align} I &= \frac {3}{3}\int \frac{\mathrm{d}x}{3x+2}\\ &= \frac{1}{3}\int \frac{3\mathrm{d}x}{3x+2}\\ &= \frac{1}{3}\int \frac{\mathrm{d}(3x)}{3x+2}\\ &= \frac{1}{3}\int \frac{\mathrm{d}(3x+2)}{3x+2}\\ &= \frac{1}{3}\ln (3x+2) + C \end{align}

Integrate 1/2+3xdx............ let u=2+3x, so that 3dx=du, .'. dx=du/3u.......and we know........int:du/u=In\u/+C, where c is the arbitrary constant for indifinite integration........ Therefore, int:du/3u=1/3 In u+c..........final answer is...........1/3In(2+3x)+C. Hope this help........ By O'john

• Please use $\LaTeX$ formatting for your math, and please do not use "..." as spacers. – apnorton Jan 17 '14 at 16:25