Question

evaluate the following integrals: integral[1/9x^2+6x+5]DX​

Answer 1

$\implies \displaystyle \int ( \dfrac{1}{9} {x}^{2} + 6x + 5)dx$

$\implies \displaystyle \int \dfrac{1}{9} {x}^{2} \: dx+\displaystyle \int 6x \: dx +\displaystyle \int 5dx$

$\implies \dfrac{1}{9}\displaystyle \int {x}^{2} \: dx+6\displaystyle \int x \: dx +5\displaystyle \int 1 \: \:dx$

$\implies( \dfrac{1}{9} \times \dfrac{ {x}^{3}}{3} \: )+(6 \times \dfrac{{x}^{2} }{2} ) +5x + c$

$\implies \dfrac{ {x}^{3}}{27} \:+(3\times \dfrac{{x}^{2} }{1} ) +5x + c$

$\implies \dfrac{ {x}^{3}}{27} \:+3{x}^{2}+5x$

Answer :

$\dfrac{ {x}^{3}}{27} \:+3{x}^{2}+5x$

___________________

Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

___________________