Q. 24 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \left(\frac{2}{x-1}-\frac{3}{2x+1}\right)\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \left(\frac{2}{\mathrm{x}-1}-\frac{3}{2\mathrm{x}+1}\right)\mathrm{dx}=2\log (\mathrm{x}-1)-\frac{3}{2}\log (2\mathrm{x}+1)+C$.

Step 1. Given Information

Solving the given integrals.
$\int \left(\frac{2}{x-1}-\frac{3}{2x+1}\right)\mathrm{d}x$

Step 2. Solving the given integral using substitution method.

Let

$$\begin{gathered} u=x-1 v=2x+1 \\ \frac{du}{dx}=1 \frac{dv}{dx}=2 \\ du=dx dv=2dx \\ du=dx \frac{1}{2}dv=dx \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \left(\frac{2}{x-1}-\frac{3}{2x+1}\right)\mathrm{d}x=\int \frac{2}{x-1}\mathrm{d}x-\int \frac{3}{2x+1}\mathrm{d}x \\ \int \left(\frac{2}{x-1}-\frac{3}{2x+1}\right)\mathrm{d}x=2\int \frac{1}{u}\mathrm{du}-\frac{3}{2}\int \frac{1}{v}\mathrm{dv} \\ \int \left(\frac{2}{\mathrm{x}-1}-\frac{3}{2\mathrm{x}+1}\right)\mathrm{dx}=2\mathrm{logu}-\frac{3}{2}\mathrm{logv}+\mathrm{C} \\ \int \left(\frac{2}{\mathrm{x}-1}-\frac{3}{2\mathrm{x}+1}\right)\mathrm{dx}=2\log (\mathrm{x}-1)-\frac{3}{2}\log (2\mathrm{x}+1)+\mathrm{C} \end{gathered}$$