6BLinearSubstitution

Linear Substitution

Sometimes a linear term appears within a more complex expression. A substitution approach is often still useful because the derivative of a linear term is a constant. An additional power of x in the integral can be expressed in terms of u.

Note: Maths Methods teaches a rule to use for powers of linear terms, but this method is better since it leads to understanding rather than blind application of a rule (and the method has more widespread use so is more useful).

Example 1

math . \qquad \displaystyle{ \int { (3x-1)^4 } \; dx } math

3x – 1 is within a more complex expression

math \\ . \qquad \text {Let } \; u = 3x - 1 \\. \\ . \qquad \text {so } \; \dfrac {du}{dx} = 3 \\. \\ . \qquad \text {so } \; dx = \dfrac {du}{3} math

Substitute **u** and **dx** into the integral

math \\ . \qquad \displaystyle{ \int { (3x-1)^4 } \; dx} \\. \\ . \qquad = \displaystyle{ \int { u^4 } \; \dfrac {du}{3} } \\. \\ . \qquad = \dfrac {1}{3} \displaystyle{ \int { u^4 } \; du} math . math \\ . \qquad = \dfrac {u^5}{15} + c \\. \\ . \qquad = \dfrac {(3x-1)^5}{15} + c math

Example 2

math . \qquad \displaystyle{ \int{ 3x \sqrt{2x+3} } \; dx} math

2x + 3 is in a more complex expression and 3x is NOT the derivative.

math \\ . \qquad \text{Let } \; u=2x+3 \\. \\ . \qquad \text {so } \; \dfrac {du}{dx} = 2 \\. \\ . \qquad \text {so } \; dx = \dfrac {du}{2} math . math . \qquad \text{Also } \; x = \dfrac {u-3}{2} math

Substitute u and dx into the integral

math \\ . \qquad \displaystyle{ \int { 3x \sqrt{2x+3} } \;dx } \\. \\ . \qquad = \displaystyle{ \int{ 3 \left( \dfrac{u-3}{2} \right) \sqrt{u} \; \dfrac {du}{2} } } \\. \\ . \qquad = \dfrac{3}{4} \displaystyle{ \int{ \left( u-3 \right) u^{\frac{1}{2}} } \; du} math . math . \qquad = \dfrac{3}{4} \displaystyle{ \int{ u^{\frac{3}{2}} - 3u^{\frac{1}{2}} } \; du} math . math . \qquad = \dfrac{3}{4} \left( \dfrac { u^{\frac{5}{2}} }{\frac{5}{2} } - \dfrac { 3u^{\frac{3}{2}} }{\frac{3}{2} } \right) + c math . math . \qquad = \dfrac{3}{4} \left( \dfrac { 2u^{\frac{5}{2}} }{5} - \dfrac { 6u^{\frac{3}{2}} }{3} \right) + c math . math . \qquad = \dfrac { 3u^{\frac{5}{2}} }{10} - \dfrac { 3u^{\frac{3}{2}} }{2} + c math . math . \qquad = \dfrac { 3(2x+3)^{\frac{5}{2}} }{10} - \dfrac { 3(2x+3)^{\frac{3}{2}} }{2} + c math

math \\ . \qquad \text{Because of } \sqrt{2x+3} \; \text{ in the original integral (and in the answer): } \\ .\\ . \qquad 2x+3 \geqslant 0 \; \text{ so } x \geqslant -\dfrac{3}{2} math .
 * NOTE: **