Integral Exercise 24
We are asked to evaluate the following integral:
$$ \int \frac{9x-3}{x^2+1} \ dx $$
We begin by splitting the integrand into two simpler fractions:
$$ \int \frac{9x}{x^2+1} - \frac{3}{x^2+1} \ dx $$
Using the linearity of integration, we can rewrite this as the difference of two integrals:
$$ \int \frac{9x}{x^2+1} \ dx - \int \frac{3}{x^2+1} \ dx $$
$$ 9 \int \frac{x}{x^2+1} \ dx - 3 \int \frac{1}{x^2+1} \ dx $$
The second integral is straightforward, since ∫ 1 / (x2 + 1) = arctan(x) + c
$$ 9 \int \frac{x}{x^2+1} \ dx - 3 \arctan(x) + c $$
To solve the first integral, we’ll use a substitution. Let t = x2 + 1
$$ t = x^2 + 1 $$
$$ dt = 2x \, dx $$
Solving for dx gives:
$$ dx = \frac{1}{2x} \, dt $$
We now substitute dx = 1/(2x) dt into the integral:
$$ 9 \int \frac{x}{x^2+1} \cdot \frac{1}{2x} \ dt - 3 \arctan(x) + c $$
$$ 9 \int \frac{1}{x^2+1} \cdot \frac{1}{2} \ dt - 3 \arctan(x) + c $$
$$ \frac{9}{2} \int \frac{1}{x^2+1} \ dt - 3 \arctan(x) + c $$
Recalling that t = x2 + 1, the integral becomes:
$$ \frac{9}{2} \int \frac{1}{t} \ dt - 3 \arctan(x) + c $$
This is another standard result: ∫ 1/t = log|t| + c
$$ \frac{9}{2} \log|t| - 3 \arctan(x) + c $$
Substituting back t = x^2 + 1, we obtain:
$$ \frac{9}{2} \log|x^2 + 1| - 3 \arctan(x) + c $$
Since \( x^2 + 1 \) is always positive for all real values of \( x \), we can safely drop the absolute value:
$$ \frac{9}{2} \log(x^2 + 1) - 3 \arctan(x) + c $$
And that’s our final answer.
Done and dusted.
