Understanding Indefinite Integral of Rational Functions
Using Indefinite Integral of Rational Functions requires that the format of the expression must be power of linear expressions, such as \( (3x-1)^3, (2x+3)^3, \sqrt{4x-1} \), etc. \( (3x^2-1)^3, (2\sqrt{x}+3)^3, \sqrt{4x^3-1} \) are not applicable for this formula.
$$\displaystyle \int{(ax+b)^n}dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C \ (n \ne -1)$$
Practice Questions
Question 1
Find \( \displaystyle \int{(2x-1)^3}dx \).
\( \begin{aligned} \displaystyle
\int{(2x-1)^3}dx &= \dfrac{(2x-1)^{3+1}}{2(3+1)} + C \\
&= \frac{1}{8}(2x-1)^4 + C \\
\end{aligned} \\ \)
Question 2
Find \( \displaystyle \int{\sqrt{3x+2}}dx \).
\( \begin{aligned} \displaystyle \require{color}
\int{\sqrt{3x+2}}dx &= \int{(3x+2)^{\frac{1}{2}}}dx &\color{red} \text{convert to index form} \\
&= \dfrac{(3x+2)^{\frac{1}{2}+1}}{3\big(\frac{1}{2}+1\big)} + C \\
&= \dfrac{(3x+2)^{\frac{3}{2}}}{\frac{9}{2}} + C \\
&= \frac{2}{9}\sqrt{(3x+2)^3} + C &\color{red} \text{don’t forget to convert back to radical form} \\
\end{aligned} \\ \)
Question 3
Find \( \displaystyle \int{\frac{1}{(5x+4)^2}}dx \).
\( \begin{aligned} \displaystyle \require{color}
\int{\frac{1}{(5x+4)^2}}dx &= \int{(5x+4)^{-2}}dx &\color{red} \text{convert to index form} \\
&= \dfrac{(5x+4)^{-2+1}}{5(-2+1)} + C \\
&= -\dfrac{(5x+4)^{-1}}{5} + C \\
&= -\frac{1}{5(5x+4)} + C &\color{red} \text{convert to positive index} \\
\end{aligned} \\ \)
Question 4
Find \( \displaystyle \int{\frac{1}{\sqrt{6x-1}}}dx \).
\( \begin{aligned} \displaystyle \require{color}
\int{\frac{1}{\sqrt{6x-1}}}dx &= \int{(6x-1)^{-\frac{1}{2}}}dx &\color{red} \text{convert to index form} \\
&= \dfrac{(6x-1)^{-\frac{1}{2}+1}}{6(-\frac{1}{2}+1)} + C \\
&= \dfrac{(6x-1)^{\frac{1}{2}}}{3} + C \\
&= \frac{1}{3}\sqrt{6x-1} + C &\color{red} \text{convert back to radical form} \\
\end{aligned} \\ \)
Question 5
Find \( \displaystyle \int{\sqrt[3]{(2-3x)^4}}dx \).
\( \begin{aligned} \displaystyle \require{color}
\int{\sqrt[3]{(2-3x)^4}}dx &= \int{(2-3x)^{\frac{4}{3}}}dx &\color{red} \text{convert to index form} \\
&= \dfrac{(2-3x)^{\frac{4}{3}+1}}{-3\big(\frac{4}{3}+1\big)} + C \\
&= \dfrac{(2-3x)^{\frac{7}{3}}}{-7} + C \\
&= -\frac{1}{7} \sqrt[3]{(2-3x)^{7}} + C &\color{red} \text{convert back to radical form} \\
\end{aligned} \\ \)
