Question

If you have 1/x then the antiderevative is ln(x). Does this mean if you have 1/(1+x^2) you could do either arctan(x) or ln(1+x^2)?

Answer 1

nope you cant........

Answer 2

it does not make sense to me because they look almost the same

Answer 3

arctan x is correct the ln is not

Answer 4

you can extend the standard integrals if you replace x by a linear function of x but but not with a quadratic function

eg for antiderivative of 1 / 2x + 1 the answer is (1/2) ln 2x + 1

Answer 5

the derivative of ln(1+x^2) = 1/(1+x^2) * 2x (as you need the chain rule)

Answer 6

ahhhh ok so if i did a substitution to make 1+x^2 = u, then i could make it ln(u)?

Answer 7

The derivative of \(\ln [f(x)]=\frac{f'(x)}{f(x)}\). In the first case we have 1 over x, and 1 is the derivative of x, so its anti-derivative is ln(x). This is not the case with \(\frac{1}{1+x^2}\). That would be true if you have \(\frac{2x}{1+x^2}\).

Answer 8

Not quite, as if you let 1+x^2 = u then du= 2x dx. So you would have that pesky 2x hanging around. The integral of 1/(1+x^2) is arctan(x) + c, to prove it you would use trig sub.

Answer 9

I meant that the anti-derivative of 2x/(1+x^2) is ln(1+x^2).

Answer 10

ok ty that makes a little more sense. i have all the trig sub antiderevive rules in my notes so ill go look at them again. ty for you help i have always wondered about this

Answer 11

+c AnwarA! +c for the indefinite integral ;)

Answer 12

Yeah :D