How do you integrate 1 over x, and 1 over a linear expression?

The integral of one over x is the natural logarithm of the absolute value of x, plus C. For a linear inner expression, the integral of one over a x plus b is one over a times the natural logarithm of the absolute value of a x plus b, plus C. You must keep the factor one over a.

How do you integrate a fraction like x squared plus 1 over x?

There is no quotient rule for integration, so simplify first. Split x squared plus 1 over x into x plus one over x, then integrate term by term to get x squared over 2 plus 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. This gives the one particular anti-derivative that passes through the point.

Why must you keep the one over a when integrating 1 over a x plus b?

Because differentiating the natural logarithm of a x plus b gives a over a x plus b by the chain rule, which is a times too big. Dividing by a corrects this, so the integral of one over a x plus b is one over a times the natural logarithm of the absolute value of a x plus b, plus C.

How do you integrate 1 over x, and 1 over a linear expression?

The integral of one over x is the natural logarithm of the absolute value of x, plus C. For a linear inner expression, the integral of one over a x plus b is one over a times the natural logarithm of the absolute value of a x plus b, plus C. You must keep the factor one over a.

How do you integrate a fraction like x squared plus 1 over x?

There is no quotient rule for integration, so simplify first. Split x squared plus 1 over x into x plus one over x, then integrate term by term to get x squared over 2 plus 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. This gives the one particular anti-derivative that passes through the point.

Why must you keep the one over a when integrating 1 over a x plus b?

Because differentiating the natural logarithm of a x plus b gives a over a x plus b by the chain rule, which is a times too big. Dividing by a corrects this, so the integral of one over a x plus b is one over a times the natural logarithm of the absolute value of a x plus b, plus C.

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

Further anti-differentiation techniques

Q: What does it mean to reverse a given derivative?

Anti-differentiation reverses differentiation, so if a derivative is supplied or found with a chain, product or quotient rule, it can be read backwards to give an integral. For example, since the derivative of e to the x squared is 2 x times e to the x squared, the integral of x times e to the x squared is one half of e to the x squared, plus C.

Q: How do you combine several standard forms in one integral?

Use linearity: the integral of a sum is the sum of the integrals, and constant multiples come out the front. Integrate each term with its standard form and write a single constant of integration plus C at the end.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), further anti-differentiation combines the standard integral forms. Use linearity to integrate term by term, the reciprocal forms \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\) and \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\), simplify an integrand until a standard form appears, reverse a given derivative, and use a boundary point to fix \(+C\).

Anti-differentiation (finding the indefinite integral) is the reverse of differentiation. Once each standard form is known — the power rule \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) (\(n\neq-1\)), \(\displaystyle\int e^{x}\,dx=e^{x}+C\), \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), and \(\displaystyle\int \sin x\,dx=-\cos x+C\), \(\displaystyle\int \cos x\,dx=\sin x+C\) — the ``further'' techniques put them to work together.

By linearity, the integral of a sum is the sum of the integrals and constant multiples come out the front, so several forms are handled in one integral. The reciprocal forms extend to a linear inner: \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\), where the factor \(\dfrac{1}{a}\) must be kept.

Because there is no product or quotient rule for integration, you often simplify first — split a fraction term by term, expand a product, or write a root as a power — until a standard form appears. And since anti-differentiation reverses differentiation, a given derivative can be read backwards to produce an integral. A single boundary point then fixes the constant of integration \(+C\).

Key idea. Combine the standard forms by linearity; integrate reciprocals with \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\); simplify before integrating; reverse a given derivative; and use a boundary point to find \(+C\).

The anti-derivatives of \(\dfrac{1}{x}\) are \(y=\ln x+C\) — a family stacked by \(C\)

A boundary point picks one member: \(f(x)=3e^{x}-2\ln|2x+1|-2\) through \((0,1)\)

The standard anti-derivative forms combined by linearity:

\[\int e^{x}\,dx=e^{x}+C,\quad \int \dfrac{1}{x}\,dx=\ln|x|+C,\quad \int \sin x\,dx=-\cos x+C,\quad \int \cos x\,dx=\sin x+C\]

\int \frac{1}{x} d x = \ln | x | + C

The reciprocal form with a linear inner \(ax+b\) — keep the \(\dfrac{1}{a}\):

\[\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\]

\int \frac{1}{a x + b} d x = \frac{1}{a} \ln | a x + b | + C

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

Simplify first. There is no product or quotient rule for integration, so split a fraction (\(\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}\)), expand a product, or write a root as a power before applying a standard form.

How to handle a further anti-differentiation problem

Simplify the integrand. Split a fraction term by term, expand a product, or write a root or reciprocal as a power, until every term is a standard form.
Integrate term by term. Apply linearity: take constant multiples out the front and integrate each term with its standard form.
Use the right reciprocal form. \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\); for a linear inner, \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\) — keep the \(\dfrac{1}{a}\) and the absolute value.
Reverse a given derivative when offered. If a derivative is provided (or found with a chain, product or quotient rule), read it backwards to write the matching integral, dividing by any constant factor.
Add \(+C\), then fix it if a boundary point is given. Substitute the point (or an initial value) into the general anti-derivative and solve for \(C\).

Rate models. These techniques recover a quantity from its rate of change — displacement from velocity, a population from a growth rate, a cost from a marginal cost — when the rate mixes several standard forms. Integrate, then use a starting value to fix \(C\).

Example 1 — Reciprocal, linear inner

Find \(\displaystyle\int \dfrac{1}{2x+1}\,dx\).

Solution

Use \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\) with \(a=2\); keep the \(\dfrac{1}{2}\).

\(\displaystyle\int \dfrac{1}{2x+1}\,dx\)

\(=\)

\(\dfrac{1}{2}\ln|2x+1|+C\)

\frac{1}{2} \ln | 2 x + 1 | + C

Example 2 — Split a fraction

Find \(\displaystyle\int \dfrac{x^{2}+1}{x}\,dx\).

Solution

There is no quotient rule — split term by term first.

\(\dfrac{x^{2}+1}{x}\)	\(=\)	\(x+\dfrac{1}{x}\)
\(\displaystyle\int \left(x+\dfrac{1}{x}\right)dx\)	\(=\)	\(\dfrac{x^{2}}{2}+\ln\|x\|+C\)

\frac{x^{2}}{2} + \ln | x | + C

Example 3 — Reverse a given derivative

Given \(y=e^{x^{2}}\), find \(\displaystyle\int x\,e^{x^{2}}\,dx\).

Solution

Differentiate with the chain rule, then read it backwards.

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

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

Example 4 — Particular anti-derivative

\(f'(x)=3e^{x}-\dfrac{4}{2x+1}\) and \(f(0)=1\). Find \(f(x)\).

Solution

Integrate term by term (the \(\dfrac{4}{2x+1}\) term gives \(2\ln|2x+1|\)), then substitute \((0,1)\).

\(f(x)\)	\(=\)	\(3e^{x}-2\ln\|2x+1\|+C\)
\(1\)	\(=\)	\(3-0+C \Rightarrow C=-2\)
\(f(x)\)	\(=\)	\(3e^{x}-2\ln\|2x+1\|-2\)

f (x) = 3 e^{x} - 2 \ln | 2 x + 1 | - 2

Common pitfalls

Keep the \(\dfrac{1}{a}\). \(\displaystyle\int \dfrac{1}{2x+1}\,dx=\dfrac{1}{2}\ln|2x+1|+C\), not \(\ln|2x+1|+C\).

Split before integrating. \(\displaystyle\int \dfrac{x^{2}+1}{x}\,dx=\dfrac{x^{2}}{2}+\ln|x|+C\), never \(\ln|x^{2}+1|+C\) — there is no quotient rule.

Watch the trig sign. \(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\).

Never drop the \(+C\) or the absolute value. An indefinite integral is a family of functions, and the reciprocal form is \(\ln|x|\), not \(\ln x\).

Frequently asked questions

How do you integrate 1/x and 1/(ax+b)?

\(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\); for a linear inner, \(\displaystyle\int \dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\) — keep the \(\dfrac{1}{a}\).

How do you integrate a fraction like (x^2+1)/x?

Split it first: \(\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}\), then integrate term by term to \(\dfrac{x^{2}}{2}+\ln|x|+C\).

What does it mean to reverse a given derivative?

Anti-differentiation reverses differentiation, so a supplied derivative can be read backwards. Since \(\dfrac{d}{dx}e^{x^{2}}=2x\,e^{x^{2}}\), we get \(\displaystyle\int x\,e^{x^{2}}\,dx=\dfrac{1}{2}e^{x^{2}}+C\).

How do you combine several standard forms in one integral?

Use linearity: integrate each term with its standard form, take constant multiples out the front, and write a single \(+C\).

How do you find the constant of integration?

Substitute a boundary point (or an initial value) into the general anti-derivative and solve for \(C\) — this gives the particular anti-derivative.

Why keep the 1/a in the integral of 1/(ax+b)?

Differentiating \(\ln|ax+b|\) gives \(\dfrac{a}{ax+b}\) by the chain rule — \(a\) times too big — so dividing by \(a\) corrects it.

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