Q. 67 - Calculus

Question

Solve each of the integrals in Exercises 63–68, where a, b, and c are real numbers with 

$$\begin{gathered} a\ne 0 , b\ne 0 , c>1, and c\ne 0. \\ \int \frac{a}{1+{\left(bx\right)}^2}dx. \end{gathered}$$ 

Step-by-Step Solution

Answer

The value of the integral is $\frac{a}{b}{\tan }^{-1}\left(bx\right) +k.$

Step 1. Given Information.

Given is a integral:

$$\begin{gathered} \int \frac{a}{1+{\left(bx\right)}^2}dx, \\ a\ne 0 , b\ne 0 , c>1, and c\ne 0, \\ and a,b,and c are cons\tan ts. \end{gathered}$$

Step 2. Formula involved.

$\int \frac{1}{1+x^2}dx = {\tan }^{-1}\left(x\right)+c.$

Step 3. Solving the integral.

$$\begin{gathered} \int \frac{a}{1+{\left(bx\right)}^2}dx \\ let t=bx \\ dt = bdx \\ Putting in the integral we get, \\ =\frac{a}{b}\int \frac{1}{1+t^2}dt \\ =\frac{a}{b} {\tan }^{-1}\left(t\right) + k \\ = \frac{a}{b} {\tan }^{-1}\left(bx\right) + k. \end{gathered}$$