Techniques of  Integration

International Baccalaureate Higher Level Mathematics

IB HL Math

Conceptual Video: Integration by Parts

**Integration by Parts Formula Derivation:**

\(\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\)

\(u\displaystyle\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(uv) - v\frac{\mathrm{d}u}{\mathrm{d}x}\)

Integrate both sides:
\(\displaystyle\displaystyle\int u\frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = \displaystyle\displaystyle\int \frac{\mathrm{d}}{\mathrm{d}x}(uv) \, \mathrm{d}x - \displaystyle\displaystyle\int v\frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\)

\(\displaystyle\displaystyle\int u\frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \displaystyle\displaystyle\int v\frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\)

Note: Not everything can be integrated, but most (if not all) expressions we learnt can be differentiated.

**Example:** \(\displaystyle\displaystyle\int x \ln x \, \mathrm{d}x\)

Let \(u = \ln x\), \(\displaystyle\frac{\mathrm{d}v}{\mathrm{d}x} = x\)
\(\displaystyle\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\), \(v = \displaystyle\frac{x^2}{2}\)

\(= uv - \displaystyle\displaystyle\int v\frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\)
\(= \displaystyle\frac{x^2}{2}\ln x - \displaystyle\displaystyle\int \frac{1}{x} \cdot \frac{x^2}{2} \, \mathrm{d}x\)
\(= \displaystyle\frac{1}{2}x^2 \ln x - \frac{1}{2}\displaystyle\displaystyle\int x \, \mathrm{d}x\)
\(= \displaystyle\frac{1}{2}x^2 \ln x - \frac{1}{2}\left(\frac{x^2}{2}\right) + c\)
\(= \displaystyle\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + c\)

Conceptual Video: Integration by Parts

Conceptual Video: Integration by Parts

Video Solution: