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

1 answer:

3 0

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

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