Integrals of special forms and the quadratic

What you will be able to do

• write the integrals of $\dfrac{1}{x^2 \pm a^2}$ and $\dfrac{1}{\sqrt{a^2 - x^2}}$, and of the roots of $a^2 - x^2$ and $x^2 \pm a^2$;
• complete the square so that a quadratic fits a standard form;
• integrate $\dfrac{px + q}{ax^2 + bx + c}$ by splitting the top.

Why these forms need their own rules

Substitution does not turn $\dfrac{1}{x^2 + 9}$ into a simple form. There is no inner function with its derivative beside it. Yet this integral comes up so often, in physics and in statistics as well as in maths, that its answer is worth remembering. The answer is an inverse tangent.

The six standard forms

Here $a$ is a positive constant. Each result can be checked by differentiating it.

• $\displaystyle\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x - a}{x + a}\right| + C$
• $\displaystyle\int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a + x}{a - x}\right| + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C$

The first one is the key to the others. Write $x^2 + a^2 = a^2\left(1 + \dfrac{x^2}{a^2}\right)$ and put $u = \dfrac{x}{a}$. The second and third come from $\dfrac{1}{x^2 - a^2} = \dfrac{1}{2a}\left(\dfrac{1}{x - a} - \dfrac{1}{x + a}\right)$, which is the idea of the next lesson.

Three integrals of a square root

These also come up in exams. They are found by integration by parts, which is the third lesson.

• $\displaystyle\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \sqrt{x^2 + a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C$
• $\displaystyle\int \sqrt{x^2 - a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C$

The first one is the area of part of a circle: it contains the area of a triangle and of a circular sector.

A note on words

The inverse tangent is written $\tan^{-1}$ in NCERT and in many Indian books. A-Level, IB and AP texts often write $\arctan$. The inverse sine is $\sin^{-1}$ or $\arcsin$. The function $\ln\left(x + \sqrt{x^2 + a^2}\right)$ differs from $\sinh^{-1}\dfrac{x}{a}$ only by the constant $\ln a$. So A-Level and IB texts often write the answer as $\operatorname{arsinh}\dfrac{x}{a} + C$.

Completing the square

When the quadratic is not in the standard form, complete the square first.

Then put $u = x + \dfrac{b}{2a}$, so that $du = dx$. The integral now has one of the standard forms in $u$.

Worked examples

Example 1. Find $\displaystyle\int \dfrac{dx}{x^2 + 6x + 13}$.

1. Complete the square: $x^2 + 6x + 13 = (x + 3)^2 + 4$.
2. Put $u = x + 3$. The integral is $\int \dfrac{du}{u^2 + 2^2} = \dfrac{1}{2}\tan^{-1}\dfrac{u}{2} + C$.
3. Return to $x$: $\dfrac{1}{2}\tan^{-1}\dfrac{x + 3}{2} + C$.

Example 2. Find $\displaystyle\int \dfrac{dx}{\sqrt{7 - 6x - x^2}}$.

1. Complete the square: $7 - 6x - x^2 = 16 - (x + 3)^2$.
2. Put $u = x + 3$. The integral is $\int \dfrac{du}{\sqrt{4^2 - u^2}} = \sin^{-1}\dfrac{u}{4} + C$.
3. Return to $x$: $\sin^{-1}\dfrac{x + 3}{4} + C$.

Example 3. Find $\displaystyle\int \dfrac{3x + 1}{x^2 + 2x + 5}\,dx$. The top is not the derivative of the bottom, which is $2x + 2$. So split it.

1. Write $3x + 1 = \dfrac{3}{2}(2x + 2) - 2$. Check: $\dfrac{3}{2}(2x + 2) = 3x + 3$, and $3x + 3 - 2 = 3x + 1$.
2. The first part has the top equal to the derivative of the bottom: $\dfrac{3}{2}\displaystyle\int \dfrac{2x + 2}{x^2 + 2x + 5}\,dx = \dfrac{3}{2}\ln(x^2 + 2x + 5)$.
3. The second part: $-2\displaystyle\int \dfrac{dx}{(x + 1)^2 + 4} = -2 \times \dfrac{1}{2}\tan^{-1}\dfrac{x + 1}{2}$.
4. The answer is $\dfrac{3}{2}\ln(x^2 + 2x + 5) - \tan^{-1}\dfrac{x + 1}{2} + C$.

Practise with new numbers

Try it in Python

Each row of the program has an integrand and the answer that a special form gave. The program checks each answer by taking its slope.

Common mistakes

• Putting $\dfrac{1}{a^2}$ in front of the inverse tangent. The factor is $\dfrac{1}{a}$.
• Mixing up $x^2 - a^2$ and $a^2 - x^2$. The signs inside the logarithm swap. Check with a small number.
• Skipping the square. For $x^2 + 6x + 13$ the standard form needs $(x + 3)^2 + 4$.
• Ignoring a leading number. For $4x^2 + 9$, take the $4$ out first: $4x^2 + 9 = 4\left(x^2 + \dfrac{9}{4}\right)$.
• Splitting the top wrongly. Always check: the new top, multiplied out, must equal the old top.

Remember these

• $\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$ and $\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a}$.
• $\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x - a}{x + a}\right|$, and the roots of $x^2 \pm a^2$ give logarithms.
• Complete the square, then put $u = x + \dfrac{b}{2a}$.
• For $\dfrac{px + q}{ax^2 + bx + c}$, write $px + q = A(2ax + b) + B$. The first part gives a logarithm, and the second part gives one of the standard forms.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\int \dfrac{dx}{x^2 + 25}$.

2. Find $\displaystyle\int \dfrac{dx}{9 - x^2}$ for $|x| < 3$.

3. Find $\displaystyle\int \dfrac{dx}{x^2 + 2x + 10}$.

4. Find $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - 9}}$ for $x > 3$.

5. Find $\displaystyle\int \sqrt{16 - x^2}\,dx$ for $|x| \le 4$.

References

• OpenStax, Calculus Volume 1, Chapter 5, section 5.7 Integrals Resulting in Inverse Trigonometric Functions, and Calculus Volume 2, Chapter 3, section 3.3 Trigonometric Substitution.
• NCERT, Mathematics Part II for Class XII, Chapter 7, the section on integrals of some particular functions.