Basics of Indefinite Integral Formula

An indefinite Integral Formula has been made from the reverse operations of differentiation or anti-differentiation.
$ \begin{aligned} \displaystyle
\frac{d}{dx}x^3 &= 3x^2 &\Rightarrow \int{3x^2}dx &= x^3 \\
\frac{d}{dx}(x^3 + 4) &= 3x^2 &\Rightarrow \int{3x^2}dx &= x^3 + 4 \\
\frac{d}{dx}(x^3-2) &= 3x^2 &\Rightarrow \int{3x^2}dx &= x^3-2
\end{aligned} $

The general formula of indefinite integral is; $$ \large \int{x^n}dx = \frac{x^{n+1}}{n+1} + C $$

Practice Questions

Question 1

Find $ \displaystyle \int{4}dx $.

$ \begin{aligned} \displaystyle \require{color}
\int{4}dx &= \int{4 \times 1}dx \\
&= \int{4 \times x^0}dx \\
&= \frac{4x^{0+1}}{0+1} + C \\
&= 4x + C &\color{red} \text{don’t forget to have }+C
\end{aligned} $

Question 2

Find $ \displaystyle \int{2x}dx $.

$ \begin{aligned} \displaystyle
\int{2x}dx &= \frac{2x^{1+1}}{1+1} + C \\
&= x^2 +C
\end{aligned} $

Question 3

Find $ \displaystyle \int{\frac{4x^3}{3}}dx $.

$ \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\int{\frac{4x^3}{3}}dx &= \frac{4}{3} \int{x^3} dx &\color{red} \text{always a great idea to take our the coefficient} \\
&= \frac{4}{3} \times \frac{x^{3+1}}{3+1} +C &\color{red} \text{then perform the usual integration} \\
&= \frac{x^4}{3} + C
\end{aligned} $

Question 4

Find $ \displaystyle \int{(3x^2 + 2x + 4)}dx $.

$ \begin{aligned} \displaystyle
\int{(3x^2 + 2x + 4)}dx &= \frac{3x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1} + 4x +C \\
&= x^3 + x^2 + 4x + C
\end{aligned} $

Question 5

Find $ \displaystyle \int{\frac{3}{2x^2}}dx $.

$ \begin{aligned} \displaystyle \require{color}
\int{\frac{3}{2x^2}}dx &= \frac{3}{2} \int{\frac{1}{x^2}}dx \\
&= \frac{3}{2} \int{x^{-2}}dx &\color{red} \text{make it integratable} \\
&= \frac{3}{2} \times \frac{x^{-2+1}}{-2+1} + C \\
&= \frac{3}{2} \times \frac{x^{-1}}{-1} + C \\
&= -\frac{3}{2x} + C &\color{red} \text{always leave your final answer in positive index}
\end{aligned} $

Question 6

Find $ \displaystyle \int{\frac{6x^3+2x^2-3x}{x}}dx $.

$ \begin{aligned} \displaystyle \require{color}
\int{\frac{6x^3+2x^2-3x}{x}}dx &= \int{\Big(\frac{6x^3}{x} + \frac{2x^2}{x}-\frac{3x}{x}\Big)}dx &\color{red} \text{separate into single fractions} \\
&= \int{(6x^2+2x-3)}dx \\
&= \frac{6x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1}-3x + C \\
&= 2x^3 + x^2-3x + C
\end{aligned} $

Question 7

Find $ \displaystyle \int{x(x-2)}dx $.

$ \begin{aligned} \displaystyle \require{color}
\int{x(x-2)}dx &= \int{(x^2-2x)}dx &\color{red} \text{expand} \\
&= \dfrac{x^{2+1}}{2+1}-\dfrac{2x^{1+1}}{1+1} + C \\
&= \frac{x^3}{3}-x^2 + C
\end{aligned} $

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