- \(\displaystyle \int \frac {x}{x^2+9} ~dx = \color{blue} {\frac 1 2 \ln(x^2 + 9)}\)
Solution: \(u = x^2 + 9\), \(du = 2x ~dx\)
Solution: \(\frac {x}{x^2+9} ~dx = \frac {1}{2} \cdot \frac {1}{u} ~du\)
Used 2018
Solution
1. Make the substitution \(u = x^2 + 9\).Then \(du = 2x \, dx \implies x \, dx = \frac{1}{2} \, du\).
The substitution rule for indefinite integrals states that if \(u = g(x)\) is differentiable and \(F\) is an antiderivative of \(f\), then \(\int f(g(x))\, g'(x)\, dx = F(g(x)) + C\). Here \(g(x) = x^2 + 9\) is differentiable for all \(x\) with \(g'(x) = 2x\), and \(f(u) = 1/u\) with antiderivative \(F(u) = \ln|u|\), so the substitution \(u = x^2 + 9\) is valid.
2. Rewrite and evaluate the integral: \[\int \frac{x}{x^2+9} \, dx = \int \frac{1}{u} \left(\frac{1}{2} \, du\right) = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C.\]
2.1. From 1, \(x\,dx = \frac{1}{2}\,du\), so \(\frac{x}{x^2+9}\,dx = \frac{1}{u}\cdot\frac{1}{2}\,du\).
2.2. The constant factor \(\frac{1}{2}\) pulls out of the integral.
2.3. The antiderivative of \(\frac{1}{u}\) is \(\ln|u|\), so \(\frac{1}{2}\int\frac{1}{u}\,du = \frac{1}{2}\ln|u| + C\).
3. Substitute back \(u = x^2 + 9\): Since \(x^2 + 9 > 0\) for all \(x \in \mathbb{R}\), \(|u| = x^2 + 9\), so: \[\int \frac{x}{x^2+9} \, dx = \frac{1}{2} \ln(x^2+9) + C.\]
3.1. Replacing \(u\) by \(x^2 + 9\) in \(\frac{1}{2}\ln|u| + C\) gives \(\frac{1}{2}\ln(x^2+9) + C\).
3.2. The absolute value is dropped because \(x^2 + 9 > 0\) for every real \(x\).