Q. 53 - Calculus

Question

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$).

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

Step-by-Step Solution

Answer

The value of the integral is $\frac{x^2}{\ln x}+c.$

Step 1. Given Information.

Given is a integral: $\int \frac{2x\ln x-x}{{\left(\ln x\right)}^2}dx.$

Step 2. Formula invovled.

$\frac{u}{v} = \int \frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$.

Step 3. Solving the integral.

$$\begin{gathered} \int \frac{2x\ln x-x}{{\left(\ln x\right)}^2}dx \\ = \int \frac{{\displaystyle \frac{d}{dx}}\left(x^2\right)\ln x-x^2{\displaystyle \frac{d}{dx}}\left(\ln x\right)}{{\left(\ln x\right)}^2} \\ By rever\sin g the formula given in step 2 we get, \\ =\frac{x^2}{\ln x}+c. \end{gathered}$$