Techniques of  Integration

International Baccalaureate Higher Level Mathematics

IB HL Math

Conceptual Video

**Cancelling Method (u-substitute method)**

Simple case: \(\displaystyle\displaystyle\int x(x^2+1) \, \mathrm{d}x = \displaystyle\displaystyle\int x^3 + x \, \mathrm{d}x = \frac{x^4}{4} + \frac{x^2}{2} + c\)

Complex case: \(\displaystyle\displaystyle\int x(x^2+1)^{10} \, \mathrm{d}x\)
Let \(u = x^2 + 1\), \(\displaystyle\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\), \(dx = \frac{\mathrm{d}u}{2x}\)
\(= \displaystyle\displaystyle\int x \cdot u^{10} \cdot \frac{\mathrm{d}u}{2x} = \frac{1}{2}\displaystyle\displaystyle\int u^{10} \, du\) [see MF26]
\(= \displaystyle\frac{1}{22}(x^2+1)^{11} + c\)

**More Examples:**

• \(\displaystyle\displaystyle\int \sec^2 x (2\tan x + 3)^5 \, \mathrm{d}x\): let \(u = 2\tan x + 3\)
\(= \displaystyle\frac{1}{12}(2\tan x + 3)^6 + c\)

• \(\displaystyle\displaystyle\int \frac{x^2}{x^3-1} \, \mathrm{d}x\): let \(u = x^3 - 1\)
\(= \displaystyle\frac{1}{3}\displaystyle\displaystyle\int \frac{1}{u} \, du = \frac{1}{3}\ln|x^3-1| + c\)

• \(\displaystyle\displaystyle\int \frac{e^x}{1+e^x} \, \mathrm{d}x\): let \(u = 1 + e^x\)
\(= \ln|1+e^x| + c\)

Conceptual Video

Conceptual Video

Video Solution: