Question

(x+1)(6x-5) find the indefinite Integral

Answer 1

`~~~~\displaystyle \int (x+1)(6x-5)~ dx\\\\\\=\displaystyle \int (6x^2-5x+6x-5)~ dx\\\\\\=\displaystyle \int(6x^2+x-5)~ dx\\\\\\=6\displaystyle \int x^2 ~ dx +\displaystyle \int x~dx - \displaystyle \int~ 5~ dx\\\\\\=6\cdot (x^(2+1))/(2+1) + \frac{x^(1+1)}{{1+1}}- 5x+C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int x^n ~ dx= (x^(n+1))/(n+1)+C,~~ n \\eq -1\right]\\\\\\=6\cdot (x^3 )/( 3) + \frac{x^2}2 - 5x+C\\\\\\=2x^3 +(x^2 )/(2 ) - 5x +C`