What are the antiderivites of
x/ (x^2 (squareroot of x^4-1))
and
1/ (1+(x^2/4)) no idea :{
I use this page instead of looking integrals up on tables
http://integrals.wolfram.com/index.jsp?expr=x%2F%28x%5E2%28x%5E4-1%29%5E%281%2F2%29%29&random=false
here is the second one
http://integrals.wolfram.com/index.jsp?expr=1%2F+%281%2B%28x%5E2%2F4%29%29&random=false
To find the antiderivatives of the given functions, we can use basic integration techniques. Let's solve each problem step by step:
1. Antiderivative of x / (x^2 * sqrt(x^4 - 1)):
To begin, let's rewrite the expression as:
x / (x^2 * √(x^4 - 1)) = 1 / (x * √(x^4 - 1))
Now, let's perform a substitution:
Let u = x^2 - 1, so du = 2x dx
Substituting the values in the expression, we get:
∫(1 / (x * √(x^4 - 1))) dx = ∫(1 / (√u)) * (du / 2x)
After simplifying, we have:
(1/2) ∫(1 / (√u)) du
Using power rule for integration (where n ≠ -1):
(1/2) * (2 * √u) + C
= √u + C
Finally, substituting back u = x^4 - 1:
√(x^4 - 1) + C, where C is the constant of integration.
2. Antiderivative of 1 / (1 + (x^2 / 4)):
We can solve this problem by using a trigonometric substitution. Let's apply the substitution:
Let x = 2tanθ, so dx = 2sec^2θ dθ
Substituting the values in the expression, we get:
∫(1 / (1 + (x^2 / 4))) dx
= ∫(1 / (1 + (4tan^2θ / 4))) * (2sec^2θ) dθ
= ∫(1 / (1 + tan^2θ)) * (2sec^2θ) dθ
Since 1 + tan^2θ = sec^2θ, the expression simplifies to:
∫(1 / sec^2θ) * (2sec^2θ) dθ
= ∫2 dθ
= 2θ + C
Finally, substituting back θ = arctan(x/2):
2arctan(x/2) + C, where C is the constant of integration.
Remember to include the constant of integration, as it is required for indefinite integration.