Q. 57 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises $19–64$. Use

a graph to check your answer.

${\int }_2^4\frac{3x}{1-x^2}dx$

Step-by-Step Solution

Answer

The value of expression is $\frac{-3}{2}\ln \left(5\right)$ and the graph is

Step 1. Given information

An expression is given ${\int }_2^4\frac{3x}{1-x^2}dx$

Step 2. Simplification

Simplify the expression as,

$$\begin{gathered} {\int }_2^4\frac{3x}{1-x^2}dx \\ =3{\int }_2^4\frac{x}{1-x^2}dx \\ =3\left(\frac{-1}{2}\right){\left[\ln \left(x^2-1\right)\right]}_2^4 \\ =\frac{-3}{2}\ln \left(5\right)-\frac{3}{2}\ln \left(3\right)+\frac{3}{2}\ln \left(3\right) \\ =\frac{-3}{2}\ln \left(5\right) \end{gathered}$$

And the graph is