#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 47 Maths Textbook Solution.

$\frac{e^{2}}{2}-e$

Hint: Use indefinite integral formula and put the limits to solve this integral

Given:

$\int_{1}^{2}\left ( \frac{x-1}{x^{2}} \right )e^{x}dx$

Sol:

$\int_{1}^{2}\left(\frac{x-1}{x^{2}}\right) e^{x} d x=\int_{1}^{2}\left(\frac{x}{x^{2}}-\frac{1}{x^{2}}\right)e^{x}d x$

$=\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x$

$=\int_{1}^{2}\left(\frac{1}{x} e^{x}-\frac{1}{x^{2}} e^{x}\right) d x$

$=\int_{1}^{2} \frac{1}{x} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$

Applying integration by parts method in Ist integral, then

$\int_{1}^{2}\left(\frac{x-1}{x^{2}}\right)^{x} d x=\left[\frac{1}{x} \int e^{x} d x\right]_{1}^{2}-\int_{1}^{2}\left(\frac{d}{d x}\left(\frac{1}{x}\right) \int e^{x} d x\right) d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$

$=\left[\frac{1}{x} e^{x}\right]_{1}^{2}-\int_{1}^{2} \frac{-1}{x^{2}} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$

$=\left[\frac{1}{x} e^{x}\right]_{1}^{2}+\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$

$=\left[\frac{1}{2} \cdot e^{2}-\frac{1}{1} \cdot e^{1}\right]$

$=\frac{e^{2}}{2}-e$