Question

Integration: Calculate the following integrals.

a.$\int \frac{1}{{\left(x-1\right)}^2}dx$b.$\int \frac{x-1}{x^2-1}dx$

c.$\int \frac{1}{x^2+1}dx$d.$\int \frac{x}{x^2+1}dx$

e.$\int \frac{x^2+1}{x^3}dx$f.$\int \frac{x+1}{x^2+1}dx$

Short answer

Part (a)$\int \frac{1}{{\left(x-1\right)}^2}dx=-\frac{1}{x-1}+C$

Part (b)$\int \frac{x-1}{x^2-1}dx=\ln |x+1|+C$

Part (c)$\int \frac{1}{x^2+1}dx={\tan }^{-1}x+C$

Part (d)$\int \frac{x}{x^2+1}dx=\frac{\ln |x^2+1|}{2}+C$

Part (e)$\int \frac{x^2+1}{x^3}dx=\ln \left|x\right|-\frac{1}{2x^2}+C$

Part (f)$\int \frac{x+1}{x^2+1}dx=\frac{\ln \left|x^2+1\right|}{2}+{\tan }^{-1}x+C$

Step-by-step solution

Part (a) Step 1. Calculating the integral

Given integral,$\int \frac{1}{{\left(x-1\right)}^2}dx$

$$\begin{gathered} =\frac{{\left(x-1\right)}^{-2+1}}{(-2+1)} \\ =-\frac{1}{x-1}+C \end{gathered}$$

Part (b) Step 1. Calculating the integral

Given integral, $\int \frac{x-1}{x^2-1}dx$

$$\begin{gathered} =\int \frac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}dx \\ =\int \frac{1}{\left(x+1\right)}dx \\ =\ln |x+1|+C \end{gathered}$$

Part (c) Step 1. Calculating the integral

Given integral, $\int \frac{1}{x^2+1}dx$

$={\tan }^{-1}x+C$

Part (d) Step 1. Calculating the integral

Given integral, $\int \frac{x}{x^2+1}dx$

$$\begin{gathered} =\int \frac{2x}{2\left(x^2+1\right)}dx \\ =\frac{\ln |x^2+1|}{2}+C \end{gathered}$$

Part (e) Step 1. Calculating the integral

Given integral, $\int \frac{x^2+1}{x^3}dx$

$$\begin{gathered} =\int \frac{x^2}{x^3}+\frac{1}{x^3}dx \\ =\int \frac{1}{x}+\frac{1}{x^3}dx \\ =\ln \left|x\right|+\frac{x^{-3+1}}{(-3+1)}+C \\ =\ln \left|x\right|-\frac{1}{2x^2}+C \end{gathered}$$

Part (f) Step 1. Calculating the integral

Given integral, $\int \frac{x+1}{x^2+1}dx$

$$\begin{gathered} =\int \left(\frac{x}{x^2+1}+\frac{1}{x^2+1}\right)dx \\ =\int \left(\frac{2x}{2\left(x^2+1\right)}+\frac{1}{x^2+1}\right)dx \\ =\frac{\ln \left|x^2+1\right|}{2}+{\tan }^{-1}x+C \end{gathered}$$