Question

integral[(1/109x)-1/(109x)^2 ]dx​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \bigg(\dfrac{1}{109x} \ - \ \dfrac{1}{109x^{2} } \bigg)dx$

As we know that,

Take 1/109 as common from equation, we get.

$\sf \implies \displaystyle \dfrac{1}{109} \int \bigg(\dfrac{1}{x} \ - \ \dfrac{1}{x^{2} } \bigg)dx$

$\sf \implies \displaystyle \dfrac{1}{109} \int \bigg(\dfrac{x - 1}{x^{2} } \bigg)dx$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ \int \bigg(\dfrac{x}{x^{2} }\bigg) dx \ - \ \int \bigg(\dfrac{1}{x^{2} } \bigg)dx \bigg]$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ \int \dfrac{dx}{x} \ - \ \int x^{-2} dx \bigg]$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ ln|x| \ - \ \dfrac{x^{-2 + 1} }{-2 + 1} \bigg] + C$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ ln|x| \ - \ \dfrac{x^{-1} }{-1} \bigg] + C$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ ln|x| \ - \bigg(\dfrac{-1}{x} \bigg) \bigg] + C$

$\sf \implies \displaystyle \dfrac{1}{109} \bigg[ ln|x| \ + \ \dfrac{1}{x} \bigg] + C$

                                                                                                                       

MORE INFORMATION.

Basic theorems on integration.

$\sf \implies \displaystyle \int k f(x)dx \ = k \int f(x) + c$

$\sf \implies \displaystyle \int [f(x) \pm g(x)]dx \ = \int f(x)dx \ \pm \ \int g(x)dx \ + C$

$\sf \implies \displaystyle \int \ = \dfrac{d[ f(x)dx]}{dx} \ = f(x)$

$\sf \implies \displaystyle \int \dfrac{d[f(x)dx]}{dx} \ = f(x)$