calculus problems?

Question

Find the integral (inverse trigonometric functions):

1. ∫[t/(t^4+16)]dt

Find the indefinite integral (natural logarithmic functions)

2. ∫[x(x+2)/(x^3+3x^2-4)]dx

Answer 1

#1: Okay, first we shall make the substitution θ = arctan (t²/4). This gives us 4 tan θ = t², so 4 sec² θ dθ = 2t dt, so we have:

∫2/((4 tan θ)² + 16) sec² θ dθ
∫2/(16 tan² θ + 16) sec² θ dθ
∫2/(16 sec² θ) sec² θ dθ
∫1/8 dθ
θ/8 + C
1/8 arctan (t²/4) + C

#2: First, we notice that -2 is a root of x³+3x²-4. So x+2 is a factor, and we can divide it out using synthetic division:

-2| 1 3 ..0 -4
....... -2 -2 .4
-----------------
..... 1 1 -2 0

So x³+3x²-4 = (x+2)(x²+x-2). So we have:

∫x(x+2)/((x+2)(x²+x-2)) dx
∫x/(x²+x-2) dx

However, x²+x-2 factors easily as (x+2)(x-1), so we have:

∫x/((x+2)(x-1)) dx

Here, we need a partial fraction decomposition. So set it up:

x/((x+2)(x-1)) = A/(x+2) + B/(x-1)
x = (x-1)A + (x+2)B

Now substituting x=-2 yields:

-2 = -3A, A = 2/3

And x=1 yields:

1=3B, B=1/3.

So we have the integral:

1/3 ∫2/(x+2) + 1/(x-1) dx

And now we have only linear factors, so the integration is easy:

2/3 ln |x+2| + 1/3 ln |x-1| + C

And we are done.