Indefinite Integral of Rational Functions

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 \( \displaystyle \int{\sqrt[3]{(2-3x)^4}}dx \).

\( \begin{aligned} \displaystyle \require{color}
\int{\sqrt[3]{(2-3x)^4}}dx &= \int{(2-3x)^{\frac{4}{3}}}dx &\color{red} \text{convert to index form} \\
&= \dfrac{(2-3x)^{\frac{4}{3}+1}}{-3\big(\frac{4}{3}+1\big)} + C \\
&= \dfrac{(2-3x)^{\frac{7}{3}}}{-7} + C \\
&= -\frac{1}{7} \sqrt[3]{(2-3x)^{7}} + C &\color{red} \text{convert back to radical form} \\
\end{aligned} \\ \)

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