Question Statement
Evaluate the integral:
β«4+x2x2β,dx
Background and Explanation
To solve this integral, we can use the technique of splitting the fraction into more manageable parts. This involves expressing the numerator in a way that simplifies the integral. In this case, we will rewrite x2 as (4+x2)β4 to split the expression into two terms. Recognizing the arctangent form for integrals will also be helpful.
Solution
- Rewrite the integrand:
Start by splitting the numerator x2 as follows:
x2=(4+x2)β4
This gives us:
4+x2x2β=4+x2(4+x2)β4β=1β4+x24β
So, the integral becomes:
β«4+x2x2β,dx=β«(1β4+x24β),dx
- Split the integral:
Now, we can break the integral into two separate parts:
β«(1β4+x24β),dx=β«1,dxβ4β«4+x21β,dx
- Solve the integrals:
- The first integral is straightforward:
β«1,dx=x
- The second integral is a standard arctangent integral. We recognize that:
β«4+x21β,dx=21βtanβ1(2xβ)
Substituting this back into our expression:
xβ4[21βtanβ1(2xβ)]
Simplifying:
xβ2tanβ1(2xβ)+C
- Splitting the Fraction: The technique of breaking the integrand into simpler parts, i.e., rewriting x2 as (4+x2)β4.
- Standard Integral of Arctangent: The formula:
β«a2+x2dxβ=a1βtanβ1(axβ)+C
Summary of Steps
-
Rewrite the integrand:
4+x2x2β=1β4+x24β
-
Split the integral into two parts:
β«1,dxβ4β«4+x21β,dx
-
Solve the integrals:
- β«1,dx=x
- β«4+x21β,dx=21βtanβ1(2xβ)
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Combine the results:
xβ2tanβ1(2xβ)+C