# trouble in solving $\int\frac{1}{t(t+2)} dt$ by using a specific variable substitution

Before anything else, I would like to apologize for my english as it is not my mother tongue and I may use it unperfectly.

I would like to solve $$\int_a^b\frac{1}{t(t+2)}dt$$. Of course, it is doable by $$\frac{1}{t(t+2)} = \frac{1}{2t} + \frac{1}{2t+2}$$ and then integrating. But I was hoping to solve it by using substitution.

I know the formula to be $$\int_a^b f'(g(x))g'(x)dx = \int_{g(a)}^{g(b)}f(x)dx$$

it seems to me that I can apply the formula here. Let : $$f(x) = ln(x) \qquad f'(x) = \frac1x$$ $$g(x) = \frac{2}{x}+1 \qquad g'(x) = -\frac{2}{x^2}$$

Therefore $$\int_a^b\frac{1}{x(x+2)}dx = \int_a^b\frac{1}{x^2(1+\frac{2}{x})}dx$$ $$\int_a^b\frac{1}{x(x+2)}dx = -\frac{1}{2}\int f'(g(x))g'(x)dx$$

I then apply the formula : $$-\frac{1}{2}\int_a^b f'(g(x))g'(x)dx = -\frac{1}{2}\int_{g(a)}^{g(b)}f(x)dx$$

$$\int_a^b\frac{1}{x(x+2)}dx = -\frac{1}{2}\int_{\frac{2}{a}+1}^{\frac{2}{b}+1}ln(x)dx$$

But those two ($$\int_a^b\frac{1}{x(x+2)}dx$$ and $$-\frac{1}{2}\int_{\frac{2}{a}+1}^{\frac{2}{b}+1}ln(x)dx$$) gives widly different result for a and b chosen randomly (for instance for (a,b) = (1,2), I get 0.202 = 0.454). Please, could someone be so kind to help me figure out what I did wrong ?

• Partial factions is a little off. Looks like you already integrated but still have integral sign Apr 16, 2019 at 18:01
• @randomgirl indeed you're right $\frac{1}{t(t+2)} = \frac{1}{2t} - \frac{1}{2t+2}$. I'll change that in my question. But my point is on computing $\int\frac{1}{x(x+2)}$ and the other directly Apr 16, 2019 at 18:17

It's because the formula should be $$\sf{\int_a^b f'(g(x))g'(x)\,dx = \color{red}{\int_{g(a)}^{g(b)}f'(x)\,dx}=[f(x)]_{g(a)}^{g(b)}=f(g(b))-f(g(a))}$$ not $$\sf{\int_{g(a)}^{g(b)} f(x)\,dx}$$ on the RHS, as the substitution $$\sf{u=g(x)}$$ confirms this. Thus you have that $$\sf{\int_a^b\frac1{x(x+2)}\,dx=-\frac12\left[\ln\left(\frac2b+1\right)-\ln\left(\frac2a+1\right)\right]=\frac12\left[\ln\frac b{b+2}-\ln\frac a{a+2}\right].}$$

• thank you ! You are indeed right. It seems that I didn't know my formulas as well as I thought. I'll re-read my learning materials. Again, thank you for your time :-). Apr 16, 2019 at 18:59
• No problem. I quite like this method, though I would have just gone for partial fractions in an exam. The derivative sign is often very easy to miss out! Apr 16, 2019 at 19:00

$$\int \frac{1}{t(t+2)} \mathrm{d} t$$

This integral could be solved using partial fraction decomposition, but there is a simpler way.

Expand fraction by $$\frac{1}{t^{2}}$$

$$=\int \frac{1}{\left(\frac{2}{t}+1\right) t^{2}} d t$$

Substitute $$u=\frac{2}{t}+1 \longrightarrow \mathrm{d} t=-\frac{t^{2}}{2} \mathrm{d} u$$

$$=-\frac{1}{2} \int \frac{1}{u} \mathrm{d} u$$

\begin{aligned} \text { Now solving: } \\ & \int \frac{1}{u} \mathrm{d} u \\ \text { This is a standard integral: } \\ &=\ln (u) \end{aligned}

\begin{aligned} \text { Plug in solved integrals: } \\ &-\frac{1}{2} \int \frac{1}{u} \mathrm{d} u \\ &=-\frac{\ln (u)}{2} \end{aligned}

Undo substitution $$u=\frac{2}{t}+1 :$$ $$=-\frac{\ln \left(\frac{2}{t}+1\right)}{2}$$

The problem is solved. Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain:

\begin{aligned} & \int \frac{1}{t(t+2)} \mathrm{d} t \\=-& \frac{\ln \left(\left|\frac{2}{t}+1\right|\right)}{2}+C \end{aligned}

• I'm sure you can do the rest , when computing a given definite integral Apr 16, 2019 at 18:37
• I also know of integral-calculator.com ;-). The way the do substitution always left me confused. I'd rather use a proven formula. I don't even know if that is a "correct" way to do so. But thank you anyway for your time. Apr 16, 2019 at 19:03
• It is correct , I just didnt want to waste time doing it myself. What they did is they used a clever step $\int \frac{1}{\left(\frac{2}{t}+1\right) t^{2}} d t$ Apr 16, 2019 at 20:02