The anti-derivative of (ax+b)^r
In Year 12 Mathematical Methods (Queensland, QCAA), a linear expression raised to a power is anti-differentiated with \(\displaystyle\int (ax+b)^{r}\,dx=\dfrac{(ax+b)^{r+1}}{a\,(r+1)}+C\) for \(r\neq-1\). You raise the index by \(1\), then divide by the new index and by the coefficient \(a\) of \(x\), and add \(+C\). This handles negative and fractional powers and surds, evaluates definite integrals, and — with a boundary point — gives a particular anti-derivative.
Anti-differentiation is the reverse of differentiation: given \(f(x)\), find \(F(x)\) with \(F'(x)=f(x)\), written \(\displaystyle\int f(x)\,dx\). Because the derivative of a constant is \(0\), the answer is a whole family \(F(x)+C\), so an indefinite integral always carries a constant of integration \(+C\).
When the function is a linear expression \(ax+b\) raised to a power \(r\), the power rule combines with reversing the chain rule. Differentiating \((ax+b)^{r+1}\) gives \((r+1)\,a\,(ax+b)^{r}\), so anti-differentiating must divide by both the new index \((r+1)\) and the coefficient \(a\):
A negative power (a reciprocal like \(\dfrac{1}{(2x-5)^{3}}=(2x-5)^{-3}\)) or a fractional power (a surd like \(\sqrt{4x+1}=(4x+1)^{1/2}\)) is first rewritten as a power of the linear expression, then the rule is applied and simplified back.
The anti-derivative of a linear expression raised to a power (\(r\) any rational number, \(r\neq-1\)):
Rewrite surds and reciprocals of the linear expression as powers first:
A definite integral uses the same anti-derivative, with no \(+C\):
How to anti-differentiate \((ax+b)^{r}\)
- Write it as a power. Turn any surd or reciprocal into an index: \(\sqrt{4x+1}=(4x+1)^{1/2}\), \(\dfrac{1}{(2x-5)^{3}}=(2x-5)^{-3}\).
- Raise the index by \(1\). The new index is \(r+1\).
- Divide by \(a\,(r+1)\). Divide by the new index and by the coefficient \(a\) of \(x\) — the extra \(\div a\) undoes the chain rule.
- Add one \(+C\) for an indefinite integral; for a definite integral, substitute the upper limit and subtract the value at the lower limit instead.
- Fix \(C\) from a boundary point. Substitute the given point (or the initial displacement for a velocity) and solve for \(C\) to get the particular anti-derivative.
Raise the index to \(5\), then divide by \(a\,(r+1)=2\times 5=10\).
| \(\displaystyle\int (2x+3)^{4}\,dx\) | \(=\) | \(\dfrac{(2x+3)^{5}}{2\times 5}+C\) |
| \(=\) | \(\dfrac{1}{10}(2x+3)^{5}+C\) |
Write \(\sqrt{4x+1}=(4x+1)^{1/2}\); divide by \(a\,(r+1)=4\times\dfrac{3}{2}=6\).
| \(\displaystyle\int (4x+1)^{1/2}\,dx\) | \(=\) | \(\dfrac{(4x+1)^{3/2}}{4\times \frac{3}{2}}+C\) |
| \(=\) | \(\dfrac{1}{6}(4x+1)^{3/2}+C\) |
Anti-differentiate (divide by \(2\times\dfrac{3}{2}=3\)), then substitute the limits.
| \(\displaystyle\int_{0}^{4}(2x+1)^{1/2}\,dx\) | \(=\) | \(\left[\dfrac{(2x+1)^{3/2}}{3}\right]_{0}^{4}\) |
| \(=\) | \(\dfrac{9^{3/2}}{3}-\dfrac{1^{3/2}}{3}=\dfrac{27-1}{3}=\dfrac{26}{3}\) |
Integrate (divide by \(2\times\dfrac{3}{2}=3\)), then substitute \((4,10)\) for \(C\).
| \(f(x)\) | \(=\) | \(\dfrac{1}{3}(2x+1)^{3/2}+C\) |
| \(10\) | \(=\) | \(\dfrac{1}{3}(9)^{3/2}+C=9+C \Rightarrow C=1\) |
| \(f(x)\) | \(=\) | \(\dfrac{1}{3}(2x+1)^{3/2}+1\) |
Common pitfalls
Frequently asked questions
How do you anti-differentiate (ax+b) to a power?
Use \(\displaystyle\int (ax+b)^{r}\,dx=\dfrac{(ax+b)^{r+1}}{a\,(r+1)}+C\) for \(r\neq-1\): raise the index by \(1\), divide by the new index and by the coefficient \(a\), add \(+C\).
Why do you divide by the coefficient of x?
Reversing the chain rule: differentiating \((ax+b)^{r+1}\) produces an extra factor \(a\), so the anti-derivative divides by \(a\) to cancel it.
How do you integrate a surd or reciprocal of a linear expression?
Rewrite as a power first: \(\sqrt{4x+1}=(4x+1)^{1/2}\), \(\dfrac{1}{(2x-5)^{3}}=(2x-5)^{-3}\); then apply the rule and simplify back.
How do you evaluate a definite integral of (ax+b) to a power?
Anti-differentiate with the rule, then substitute the upper limit and subtract the value at the lower limit; no \(+C\) is needed.
When does the rule not work?
Only at \(r=-1\), where \(r+1=0\). There \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\), a logarithm.
How do you find the constant of integration?
Substitute a given boundary point (or an initial value) into the general anti-derivative and solve for \(C\) — this gives the particular anti-derivative.