Find the most general f. Use C for the constant of the first anti-derivative and D for the constant of the second anti-derivative. f ''(x) = 4x + sin x
\[\int\limits_{}^{}f''(x) dx=f'(x)+C\] \[\int\limits_{}^{}(f'(x)+C)dx=f(x)+Cx+D\] where C and D are constants so to find f you need to find f' first
thank you for the procedure, myininaya. Do you have an answer?
\[f'(x)=4*\frac{x^{1+1}}{1+1}-cosx+C\] \[f'(x)=2x^2-cosx+C\] \[f(x)=2*\frac{x^{2+1}}{2+1}-sinx+Cx+D\]
Er...is it supposed to be "x^2+1" or did you intentionally write the numerator with "2+1" as the exponent for x?
yes 2+1 is the exponent
\[f(x)=2*\frac{x^3}{3}-sinx+Cx+D\] i was just showing you what rule i was using to integrate(showing you my work)
hipster do you got it?
Yep I got it. Thanks :)
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