Anti-derivatives?

Question

(X+2)/(x^2 +4) dx. Can anyone show me how to find the anti-derivative?

Answer 1

(x+2)/(x^2 +4)

= X/(x^2 +4) +2/(x^2 +4)

So the antiderivative of the first term is (1/2)ln|x+4|
(∫f'(x)/f(x) dx = ln|f(x)| +C)
the anti derivative of the second term is arctan(x/2)

The anti derivative of the whole lot is (1/2)ln|x+4| + arctan(x/2) + C

Answer 2

Okay, we have:

∫(x+2)/(x²+4) dx

We break this into two integrals, for reasons that will become obvious in a moment:

∫x/(x²+4) dx + ∫2/(x²+4) dx

Now for the first one, we can evaluate it using the substitution u=x²+4, du=2x dx. So it becomes:

∫1/(2u) du + ∫2/(x²+4) dx

Which is obviously:

1/2 ln |u| + ∫2/(x²+4) dx

Resubstituting:

1/2 ln |x²+4| + ∫2/(x²+4) dx

Moving our attention to the second integral, we make the substitution θ=arctan (x/2), so that x=2 tan θ. This gives dx = 2 sec² θ dθ. Then we have:

1/2 ln |x²+4| + ∫4/(4 tan² θ+4) sec² θ dθ

Canceling the fours:

1/2 ln |x²+4| + ∫1/(tan² θ+1) sec² θ dθ

We know that sec² θ = 1+tan² θ, so this is:

1/2 ln |x²+4| + ∫sec² θ/sec² θ dθ

Simplifying:

1/2 ln |x²+4| + ∫1 dθ

Which is:

1/2 ln |x²+4| + θ + C

And resubstituting:

1/2 ln |x²+4| + arctan (x/2) + C

And now we are done.

Answer 3

Let M = int[(x+2) dx / (x^2+4)]

= int[x dx / (x^2 + 4)] + int[2 dx / (x^2 + 4)]

= 0.5 int[2x dx / (x^2 + 4)] + 2 int[dx / (x^2 + 4)]

Let u = x^2 + 4, so du = 2x dx

The first half is now 0.5 int[du / u] = 0.5 ln |u| + C = 0.5 ln(x^ 2 + 4) + C

In the second one, let x = 2 tan v so dx = 2 sec^2 v dv and v = arctan(x/2).

It is now 2 int[2 sec^2 v dv / (4 + 4 tan^2 v)] = 4 int[sec^2 v dv / 4 sec^2 v] = int[dv] = v + D = arctan(x/2) + D.

So M = 0.5 ln(x^2 + 4) + arctan(x/2) + E (E = C + D)