Integrating simple algebraic expressions
Watch this video to learn about integrating algebraic expressions.
Integration is the inverse process to differentiation. Some people call it anti-differentiation.
Instead of multiplying the power at the front and subtracting one from the power, we add one to the power and then divide by the new power.
Example
\(\int {{x^2}}\,\, dx\)
Solution
This just means, integrate \({x^2}\) with respect to \(x\). Remember, add one to the power and divide by the new power.
\(\int {{x^2}}\,\, dx\)
\(= \frac{{{x^3}}}{3} + c\)
The \(+ c\) appears because when you differentiate a constant term, the answer is zero, so as we are performing 'anti-differentiation', we presume there may have been a constant term, which reduced to zero when differentiated. This \(c\) is called the constant of integration.
In general:
\(\frac{{dy}}{{dx}} = a{x^n} \to y = \frac{{a{x^{n + 1}}}}{{n + 1}} + c\) provided \(n \ne - 1\)
Question
Find \(\int {({x^4}} + {x^3})\,\,dx\)
\(\int {({x^4}} + {x^3})\,\,dx\)
\(= \frac{{{x^5}}}{5} + \frac{{{x^4}}}{4} + c\)
Question
Find \(\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx\)
\(\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx\)
\(= \frac{{4{x^4}}}{4} + \frac{{7{x^{ - 1}}}}{{ - 1}} + c\)
\(= {x^4} - \frac{7}{x} + c\)
Question
Find \(\int {{{(x + 2)}^2}}\,\,dx\)
Similar rules apply to integration whereby we need to remove the brackets first as the expression has to be sums and/or differences of terms of the form \(a{x^n}\).
\(\int {{{(x + 2)}^2}}\,\, dx\)
\(= \int {({x^2}}+ 4x + 4)\,\,dx\)
\(= \frac{{{x^3}}}{3} + \frac{{4{x^2}}}{2} + 4x + c\)
\(= \frac{{{x^3}}}{3} + 2{x^2} + 4x + c\)
Question
Find \(\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx\)
\(\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx\)
\(= \int {\frac{{x + {x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}{x}}\,\,dx\)
\(= \int {\frac{x}{x}} + \frac{{{x^{\frac{1}{2}}}}}{x} + \frac{{{x^{\frac{1}{3}}}}}{x}\,\,dx\)
\(= \int {(1 + {x^{ - \frac{1}{2}}}} + {x^{ - \frac{2}{3}}})\,\,dx\)
\(= x + \frac{{{x^{\frac{1}{2}}}}}{{\frac{1}{2}}} + \frac{{{x^{\frac{1}{3}}}}}{{\frac{1}{3}}} + c\)
\(= x + 2\sqrt x + 3\sqrt[3]{x} + c\)
