I need help on what the antiderivatives are for these 2 problems?

Question

I can't figure out these problems.

Find the general antiderivative of the following
1. (x^2 + x + 1)/(x)

Find f
2. f'(x)=2x-(3/x^4), x>0, f(1)=3

Thank you

Answer 1

For #1, there's no way you can solve that directly. Note that it can be split up into three fractions, though.

(x^2 + x + 1)/x = x^2/x + x/x + 1/x

Now, we can actually reduce that, to

x + 1 + 1/x

We solve for the antiderivative using the reverse power rule for derivatives. That is, antiderivative for x^n is [x^(n+1)]/(n+1). The exception is 1/x, where the antiderivative is ln|x|.

So the antiderivative of x+1+1/x is

(x^2) / 2 + x + ln|x|

But we want the general antiderivative, so we add a constant C.

(x^2)/2 + x + ln|x| + C

Answer 2

1. Use integration by parts, setting u=x^2 + x +1 and dv = 1/x. Do this twice and the u term should be reduced and eventually removed.

2. A basic initial value problem; use the power rule of integration to find that y = x^2 + x^(-3) + C. Use the initial value f(1) = 3 to solve for C.

Answer 3

(x^2 + x + 1)/(x)
x+1+1/x
int = x^2/2 + x +lnx +C

. f'(x)=2x-(3/x^4), x>0, f(1)=3
f(x) = x^2 +1/x^3 +C
1+1+C = 3 so C=1
f(x) = x^2+1/x^3 +1