How do you integrate a reciprocal like 1 over x cubed, or a surd like the square root of x?

Rewrite it as a power of x first. One over x cubed is x to the power minus 3, and the square root of x is x to the power one half. Then apply the power rule and simplify back. So the integral of one over x cubed is minus one over two x squared plus C, and the integral of the square root of x is two thirds x root x plus C.

Why do you always add plus C to an indefinite integral?

Because differentiating a constant gives zero, every function of the form F of x plus C has the same derivative. So an indefinite integral is a whole family of functions, and the constant of integration plus C represents all of them at once.

When does the power rule for integration not work?

The power rule fails only when n equals minus 1, because dividing by n plus 1 would mean dividing by zero. In that single case the integral of one over x is the natural logarithm of the absolute value of x, plus C.

How do you find the constant of integration from a boundary point?

Integrate to get the general anti-derivative with an unknown constant, then substitute the coordinates of the given point and solve for the constant. For example if f dashed of x is 6 x squared minus 2 and the curve passes through the point one comma four, then f of x is 2 x cubed minus 2 x plus C, and substituting gives C equals 4.

How do you integrate a reciprocal like 1 over x cubed, or a surd like the square root of x?

Rewrite it as a power of x first. One over x cubed is x to the power minus 3, and the square root of x is x to the power one half. Then apply the power rule and simplify back. So the integral of one over x cubed is minus one over two x squared plus C, and the integral of the square root of x is two thirds x root x plus C.

Why do you always add plus C to an indefinite integral?

Because differentiating a constant gives zero, every function of the form F of x plus C has the same derivative. So an indefinite integral is a whole family of functions, and the constant of integration plus C represents all of them at once.

When does the power rule for integration not work?

The power rule fails only when n equals minus 1, because dividing by n plus 1 would mean dividing by zero. In that single case the integral of one over x is the natural logarithm of the absolute value of x, plus C.

How do you find the constant of integration from a boundary point?

Integrate to get the general anti-derivative with an unknown constant, then substitute the coordinates of the given point and solve for the constant. For example if f dashed of x is 6 x squared minus 2 and the curve passes through the point one comma four, then f of x is 2 x cubed minus 2 x plus C, and substituting gives C equals 4.

Year 12 Methods (Unit 3 & 4) Anti-differentiation

Anti-differentiation of power functions

Q: What is anti-differentiation?

Anti-differentiation is the reverse of differentiation. To anti-differentiate a function f means to find a function F whose derivative is f. Because the derivative of any constant is zero, there are infinitely many such functions, differing only by a constant, so the answer is written with a constant of integration plus C.

Q: How do you anti-differentiate a power of x?

Use the power rule: the integral of x to the power n is x to the power n plus 1, divided by n plus 1, plus C, provided n is not equal to minus 1. In words, raise the index by one, divide by the new index, and add plus C. For example the integral of x to the fourth is x to the fifth over 5, plus C.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), anti-differentiation reverses differentiation. For a power of \(x\) the rule is \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\), valid for every rational index \(n\neq-1\) — so a reciprocal or a surd is integrated by first rewriting it as a power. Integrate term by term (linearity), always add the constant of integration \(+C\), and use a boundary point to find \(C\).

Anti-differentiation (or finding the indefinite integral) is the reverse of differentiation: given \(f(x)\), find a function \(F(x)\) with \(F'(x)=f(x)\). We write \(\displaystyle\int f(x)\,dx\) for the anti-derivative.

Because the derivative of any constant is \(0\), if \(F(x)\) works then so does \(F(x)+C\) for any constant \(C\). So the answer is a whole family of functions, written with a constant of integration \(+C\).

For a power function \(x^{n}\), the power rule \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) applies for every rational \(n\neq-1\). A negative index (a reciprocal like \(\dfrac{1}{x^{3}}=x^{-3}\)) or a fractional index (a surd like \(\sqrt{x}=x^{1/2}\)) must be rewritten as a power before the rule is applied, then simplified back.

Key idea. Anti-differentiate a power by \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) (\(n\neq-1\)): raise the index by \(1\), divide by the new index, add \(+C\). Rewrite reciprocals and surds as powers first.

The anti-derivatives of \(2x\) are \(y=x^{2}+C\) — a family stacked by \(C\)

A boundary point picks one member: \(f(x)=2x^{3}-2x+4\) through \((1,4)\)

The power rule for anti-differentiation (\(n\) any rational number, \(n\neq-1\)):

\[\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\qquad (n\neq-1)\]

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

Rewrite reciprocals and surds as powers before integrating:

\[\dfrac{1}{x^{n}}=x^{-n},\qquad \sqrt{x}=x^{1/2},\qquad \sqrt[3]{x}=x^{1/3},\qquad \dfrac{x^{a}}{x^{b}}=x^{a-b}\]

\frac{1}{x^{n}} = x^{- n}

Linearity — integrate term by term:

\[\int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx,\qquad \int k\,f(x)\,dx=k\int f(x)\,dx\]

\int k f (x) d x = k \int f (x) d x

The excluded case. The rule needs \(n\neq-1\), because \(n+1=0\) would divide by zero. For that one power, \(\displaystyle\int x^{-1}\,dx=\int \dfrac{1}{x}\,dx=\ln|x|+C\).

How to anti-differentiate a power function

Rewrite as powers of \(x\). Turn every reciprocal and surd into an index: \(\dfrac{1}{x^{3}}=x^{-3}\), \(\sqrt{x}=x^{1/2}\); expand products and simplify quotients (there is no product or quotient rule for integration).
Apply the power rule to each term. Add \(1\) to the index, divide by the new index: \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}\).
Add one \(+C\). Combine the terms and write a single constant of integration.
Simplify back. Rewrite negative and fractional indices as reciprocals and surds, e.g. \(x^{3/2}=x\sqrt{x}\).
Fix \(C\) if a boundary point is given. Substitute the point (or the initial displacement for a velocity) and solve for \(C\) to get the particular anti-derivative.

Rate models. Anti-differentiation recovers a quantity from its rate of change: displacement from velocity, volume from a flow rate, cost from a marginal cost. Integrate the rate, then use a known starting value to fix \(C\).

Example 1 — Negative index

Find \(\displaystyle\int \dfrac{1}{x^{3}}\,dx\).

Solution

Rewrite \(\dfrac{1}{x^{3}}=x^{-3}\), then apply the power rule (\(n=-3\neq-1\)).

\(\displaystyle\int x^{-3}\,dx\)	\(=\)	\(\dfrac{x^{-2}}{-2}+C\)
	\(=\)	\(-\dfrac{1}{2x^{2}}+C\)

- \frac{1}{2 x^{2}} + C

Example 2 — A surd

Find \(\displaystyle\int \sqrt{x}\,dx\).

Solution

Write \(\sqrt{x}=x^{1/2}\), integrate, then simplify \(x^{3/2}=x\sqrt{x}\).

\(\displaystyle\int x^{1/2}\,dx\)	\(=\)	\(\dfrac{x^{3/2}}{3/2}+C\)
	\(=\)	\(\dfrac{2}{3}x\sqrt{x}+C\)

\frac{2}{3} x \sqrt{x} + C

Example 3 — Simplify a quotient

Find \(\displaystyle\int \dfrac{x^{2}}{\sqrt{x}}\,dx\).

Solution

There is no quotient rule — simplify to a single power first.

\(\dfrac{x^{2}}{\sqrt{x}}\)	\(=\)	\(x^{2}\times x^{-1/2}=x^{3/2}\)
\(\displaystyle\int x^{3/2}\,dx\)	\(=\)	\(\dfrac{x^{5/2}}{5/2}+C\)
	\(=\)	\(\dfrac{2}{5}x^{2}\sqrt{x}+C\)

\frac{2}{5} x^{2} \sqrt{x} + C

Example 4 — Particular anti-derivative

\(f'(x)=6x^{2}-2\) and the curve passes through \((1,4)\). Find \(f(x)\).

Solution

Integrate, then substitute \((1,4)\) to find \(C\).

\(f(x)\)	\(=\)	\(2x^{3}-2x+C\)
\(4\)	\(=\)	\(2-2+C \Rightarrow C=4\)
\(f(x)\)	\(=\)	\(2x^{3}-2x+4\)

f (x) = 2 x^{3} - 2 x + 4

Common pitfalls

Divide by the new index, not the old one. \(\displaystyle\int x^{4}\,dx=\dfrac{x^{5}}{5}+C\), not \(\dfrac{x^{4}}{4}\) and never \(\dfrac{x^{5}}{4}\).

Rewrite before integrating. You cannot apply the power rule to \(\dfrac{1}{x^{3}}\) or \(\sqrt{x}\) as written — convert to \(x^{-3}\) and \(x^{1/2}\) first.

The rule fails at \(n=-1\). \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), not \(\dfrac{x^{0}}{0}\); the power rule cannot divide by zero.

Never drop the \(+C\). An indefinite integral is a family of functions; leaving off \(+C\) loses all but one of them and any boundary condition cannot be applied.

Frequently asked questions

What is anti-differentiation?

It is the reverse of differentiation: finding a function \(F\) whose derivative is the given \(f\). Since a constant differentiates to \(0\), the answer carries a constant of integration \(+C\).

How do you anti-differentiate a power of x?

Use the power rule \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) for \(n\neq-1\): raise the index by \(1\), divide by the new index, add \(+C\).

How do you integrate 1/x^3 or the square root of x?

Rewrite as powers first: \(\dfrac{1}{x^{3}}=x^{-3}\) gives \(-\dfrac{1}{2x^{2}}+C\), and \(\sqrt{x}=x^{1/2}\) gives \(\dfrac{2}{3}x\sqrt{x}+C\).

Why do you always add +C?

Because every function \(F(x)+C\) has the same derivative, so the indefinite integral is a whole family; \(+C\) represents all of them.

When does the power rule not work?

Only at \(n=-1\), where \(n+1=0\). There \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\).

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.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

← Previous subtopic

Anti-differentiation of polynomial functions

Next subtopic →

The anti-derivative of (ax+b)^r