The improper integral of $1/(x^2+x)$ from $0$ to $\infty$.

Question

I know the integral of $\dfrac{1}{x^2+x}$ from $0$ to infinity is equal to $$\lim_{t \to \infty}(\ln|t|-\ln|t+1|-\ln|1|+\ln|2|)ln|t|=\infty $$ but in my book the answer is $\ln2$.

Why??

Answer 1

It follows from $$\lim_{t \to \infty} \ln |t| - \ln|t + 1| = \lim_{t \to \infty} \ln \frac{t}{t + 1} = \ln \lim_{t \to \infty} \frac{t}{t + 1} = \ln 1 = 0$$

Answer 2

Is the integral $0$ to $\infty$ or $1$ to $\infty$?

Here is the indefinite integral:

$$ \frac{1}{x^2 + x} = \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$$

Integrating $$ \int \frac{1}{x^2 + x} dx =\ln(x) - \ln(x+1) = \ln\left(\frac{x}{x+1}\right)$$

If the limits are $1$ to $\infty$ then $$ \left. \ln\left(\frac{x}{x+1}\right)\right|_1^{\infty} = \ln(1)-\ln(1/2) = \ln 2$$

Answer 3

Notice that $\dfrac{1}{x^2+x}$ has a discontinuity at $x=0$. Hence, we need to consider $\displaystyle\int_0^\infty \dfrac{1}{x^2+x} \, \rm{d}x = \displaystyle\lim_{a \to 0^+} \int_a^1 \dfrac{1}{x^2+x} \, \rm{d}x+ \lim_{b \to \infty} \int_1^b \dfrac{1}{x^2+x} \, \rm{d}x$.

(Note, here I chose the upper limit of the first integral (on the right-hand side of the $=$ symbol) and the lower limit of the second integral (on the right-hand side of the $=$ symbol) to be $1$. As long as they are the same value and do not give a discontinuity for the integrand and are in the region you want to integrate, you can choose those however you'd like.)

Let's just look at the first integral. (Note: Below, I use $\log(x)$ to indicate the natural log.) We have that \begin{align}\displaystyle\lim_{a \to 0^+} \int_a^1 \dfrac{1}{x^2+x} \, \rm{d}x &= \lim_{a \to 0^+} \Big[\log(x)-\log(x+1)\Big]_a^1\\ &= \lim_{a \to 0^+} \Big[\log(1)-\log(2)-[\log(a)-\log(a+1)]\Big]\\ &= \lim_{a \to 0^+} [- \log(2) - \log(a)+ \log(a+1)]\\ &= - \log(2) - \lim_{a \to 0^+} \log(a) + \lim_{a \to 0^+} (a+1)\\ &= - \log(2) - (\infty) + 0\\ &= \infty. \end{align}

The second integral is $\displaystyle\lim_{b \to \infty} \int_1^b \dfrac{1}{x^2+x} \, \rm{d}x= \log(2)$.

So, we have that $\displaystyle\int_0^\infty \dfrac{1}{x^2+x} \, \rm{d}x = \displaystyle\lim_{a \to 0^+} \int_a^1 \dfrac{1}{x^2+x} \, \rm{d}x+ \lim_{b \to \infty} \int_1^b \dfrac{1}{x^2+x} \, \rm{d}x= \infty + \log(2) = \infty$.

Answer 4

The integrand behaves like $1/(x(x+1)) \approx 1/x$ for $x$ close to $0$ and you know $\int_0^1 1/x\,dx$ diverges. Actually, $\frac{1}{x^2+x}=\frac{1}{x(x+1)}\geq \frac{1}{2x}$ for $x\in (0,1]$ so $\int_0^1 \frac{1}{x^2+x}\,dx = \infty$. Since the function $\frac{1}{x^2+x}$ is positive on $[1,\infty)$, it follows that $\int_0^\infty \frac{1}{x^2+x}\,dx=\infty$. No need for partial fractions.

I suspect that you copied the problem down wrong and the lower limit of integration is supposed to be $1$.