- \(\displaystyle \int \ln ^2 (x) ~dx = \color{blue} {2 x - 2 x \ln (x) + x \ln ^2 (x)}\)
Solution: \(u = \ln ^2 (x)\), \(v = x\), \(du = \frac {2 \ln (x)}{x} ~dx\), \(dv = dx\)
Solution: \(x \ln ^2 (x) - \int x \cdot \frac {2 \ln (x)}{x} ~dx = x \ln ^2 (x) - 2(x \ln (x) - x)\)
Used 2019
\(\displaystyle \int x^a e^{bx} ~dx = \frac {1}{b} x^a e^{bx} - \int \frac {a}{b} x^{a-1} e^{bx}\)
\(\displaystyle \int xe^{2x} ~dx = \color {blue} {\frac {1}{4} (2x-1) e^{2x}}\)
\(\displaystyle \int x^3 e^{-x} ~dx = \color {blue} {- (x^3 + 3x^2 + 6x + 6) e^{-x}}\)
\(\displaystyle \int_0^{\infty} x^3 e^{-x} ~dx = \color {blue} {6}\)
\(\displaystyle \int x^2e^x ~dx =\color {blue}{(x^2 - 2x + 2) e^x}\)
- Used 2018
5. \(\displaystyle \int xe^{-x} ~dx = \color {blue} {- (x + 1) e^ {-x}}\)
6. \(\displaystyle \int_{0}^{\infty} x^3e^{-x^2} ~dx = \color {blue} {\frac {1}{2}}\)
Solution
1. \(\int (\ln x)^2 \, dx = x (\ln x)^2 - 2x \ln x + 2x + C\).
Integrate by parts, \(\int u \, dv = uv - \int v \, du\), with \(u = (\ln x)^2\) and \(dv = dx\), so \(du = \frac{2 \ln x}{x} \, dx\) and \(v = x\): \[\int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} \, dx = x (\ln x)^2 - 2 \int \ln x \, dx.\] By parts again with \(u = \ln x\), \(dv = dx\), so \(du = \frac{1}{x}\,dx\), \(v = x\): \[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C_1.\] Substituting, \[\int (\ln x)^2 \, dx = x (\ln x)^2 - 2(x \ln x - x) + C = x (\ln x)^2 - 2x \ln x + 2x + C.\]
2. For \(b \neq 0\), the reduction formula \(\int x^a e^{bx} \, dx = \frac{1}{b} x^a e^{bx} - \frac{a}{b} \int x^{a-1} e^{bx} \, dx\) holds.
By parts with \(u = x^a \implies du = a x^{a-1} dx\), and \(dv = e^{bx} dx \implies v = \frac{1}{b} e^{bx}\).
3. The following cases of the reduction formula hold.
\(\int x e^{2x} \, dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C = \frac{1}{4}(2x - 1)e^{2x} + C\).
\(\int x^3 e^{-x} \, dx = -(x^3 + 3x^2 + 6x + 6)e^{-x} + C\).
\(\int_0^\infty x^3 e^{-x} \, dx = \Gamma(4) = 3! = 6\).
With \(t = x^2\), \(dt = 2x\,dx\): \(\int_0^\infty x^3 e^{-x^2} \, dx = \frac{1}{2} \int_0^\infty t e^{-t} \, dt = \frac{1}{2} \Gamma(2) = \frac{1}{2}\).