Integral Calculation Exercise 7
We aim to evaluate the integral
$$ \int \frac{x^2}{4x^2+1} \ dx $$
To simplify the computation, we try to manipulate the integrand so that the numerator mirrors the denominator.
We start by multiplying and dividing the integrand by 4:
$$ \int \frac{4}{4} \cdot \frac{x^2}{4x^2+1} \ dx $$
$$ \int \frac{1}{4} \cdot \frac{4x^2}{4x^2+1} \ dx $$
Factoring out the constant \( \frac{1}{4} \):
$$ \frac{1}{4} \cdot \int \frac{4x^2}{4x^2+1} \ dx $$
We now rewrite the numerator by adding and subtracting 1:
$$ \frac{1}{4} \cdot \int \frac{4x^2 + 1 - 1}{4x^2+1} \ dx $$
$$ \frac{1}{4} \cdot \int \left( \frac{4x^2 + 1}{4x^2+1} - \frac{1}{4x^2+1} \right) \ dx $$
The first term simplifies to 1:
$$ \frac{1}{4} \cdot \int \left( 1 - \frac{1}{4x^2+1} \right) \ dx $$
Applying the linearity of the integral, we can split this into two separate integrals:
$$ \frac{1}{4} \cdot \left[ \int 1 \ dx - \int \frac{1}{4x^2+1} \ dx \right] $$
The first integral is immediate: \( \int 1 \ dx = x + c \), where \( c \) is an arbitrary constant of integration.
$$ \frac{1}{4} \cdot \left[ x + c - \int \frac{1}{4x^2+1} \ dx \right] $$
We rewrite the integrand in the second term using the identity \( 4x^2 = (2x)^2 \):
$$ \frac{1}{4} \cdot \left[ x + c - \int \frac{1}{(2x)^2 + 1} \ dx \right] $$
To evaluate this integral, we apply the substitution method with the auxiliary variable u = 2x:
$$ u = 2x $$
Differentiating both sides gives:
$$ du = 2 \ dx \quad \Rightarrow \quad dx = \frac{du}{2} $$
Substituting into the integral yields:
$$ \frac{1}{4} \cdot \left[ x + c - \int \frac{1}{(2x)^2 + 1} \ dx \right] $$
$$ \frac{1}{4} \cdot \left[ x + c - \int \frac{1}{u^2 + 1} \cdot \frac{du}{2} \right] $$
Extracting the constant \( \frac{1}{2} \):
$$ \frac{1}{4} \cdot \left[ x + c - \frac{1}{2} \cdot \int \frac{1}{u^2 + 1} \ du \right] $$
This is a standard integral: \( \int \frac{1}{u^2 + 1} \ du = \arctan(u) + c \).
Substituting back \( u = 2x \), we obtain:
$$ \frac{1}{4} \cdot \left[ x + c - \frac{1}{2} \cdot \arctan(2x) \right] $$
Expanding the expression gives:
$$ \frac{x}{4} - \frac{1}{8} \cdot \arctan(2x) + c $$
Therefore, the solution to the integral is:
$$ \int \frac{x^2}{4x^2+1} \ dx = \frac{x}{4} - \frac{1}{8} \cdot \arctan(2x) + c $$
And so on.
