Special integrals and radical forms

A few patterns recur so often that recognising them is part of the calculus literacy you need. Each is a derivative of a $\ln$ or an inverse-trigonometric function in disguise.

The denominators-with-squares family

The first two come from partial fractions: $\dfrac{1}{x^2 - a^2} = \dfrac{1}{2a}\left(\dfrac{1}{x-a} - \dfrac{1}{x+a}\right)$. The middle one is the definition of $\tan^{-1}$. The radical forms come from trigonometric substitution.

Completing the square

The integrals above always have a "naked" $x^2 + a^2$ or $x^2 - a^2$. Real problems give you $ax^2 + bx + c$. The strategy: complete the square to write $ax^2 + bx + c = a\left[\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a^2}\right].$ Then substitute $u = x + b/(2a)$ to reach one of the standard forms.

Linear numerator over quadratic

For $\int \dfrac{px + q}{ax^2 + bx + c}\,dx$ , algebraic or under a radical , split the numerator: write $px + q = \dfrac{p}{2a}(2ax + b) + \left(q - \dfrac{pb}{2a}\right)$. The first chunk is a perfect derivative of the denominator (giving a log or $2\sqrt{\cdot}$). The second is a constant times a standard form. This decomposition is in routine board-exam playbooks.

Two further integrals

Two more patterns worth knowing.

$\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}(x/a) + C.$

$\int \sqrt{x^2 + a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 + a^2} + \dfrac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C.$

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

Each is derived by trigonometric (or hyperbolic) substitution. Memorise the patterns; don't try to re-derive on exam day.

Derivations of two key results

Why is $\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C$?

Substitute $x = a \tan\theta$. Then $dx = a\sec^2\theta\,d\theta$, $\sqrt{x^2 + a^2} = a\sec\theta$. The integral becomes $\int \dfrac{a\sec^2\theta}{a\sec\theta}\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C$. Restore $x$: $\tan\theta = x/a$, $\sec\theta = \sqrt{x^2 + a^2}/a$, so $\sec\theta + \tan\theta = \dfrac{x + \sqrt{x^2 + a^2}}{a}$. Absorbing $\ln a$ into the constant: $\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C.$

Why does $\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}(x/a)$?

Substitute $x = a\sin\theta$. Then $\sqrt{a^2 - x^2} = a\cos\theta$, $dx = a\cos\theta\,d\theta$. Integral becomes $\int a^2 \cos^2\theta\,d\theta = a^2 \cdot \dfrac{\theta + \sin\theta\cos\theta}{2}$. Restore: $\theta = \sin^{-1}(x/a)$, $\sin\theta\cos\theta = (x/a)(\sqrt{a^2-x^2}/a)$. Multiply through and rearrange to get the stated form.

Worked examples

Example 1. $\int \dfrac{dx}{x^2 + 2x + 5}$.

Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$. So $\int \dfrac{dx}{(x+1)^2 + 2^2} = \dfrac{1}{2}\tan^{-1}\!\left(\dfrac{x+1}{2}\right) + C$.

Example 2. $\int \dfrac{dx}{\sqrt{4x^2 - 9}}$.

Factor: $\sqrt{4x^2 - 9} = 2\sqrt{x^2 - 9/4}$. So $\int \dfrac{dx}{2\sqrt{x^2 - (3/2)^2}} = \dfrac{1}{2}\ln\left|x + \sqrt{x^2 - 9/4}\right| + C$.

Example 3. $\int \dfrac{2x + 3}{\sqrt{x^2 + 4x + 5}}\,dx$.

Split: $2x + 3 = (2x + 4) - 1$. So the integral splits as $\int \dfrac{2x + 4}{\sqrt{x^2 + 4x + 5}}\,dx - \int \dfrac{dx}{\sqrt{x^2 + 4x + 5}}.$ First piece: numerator is derivative of $x^2 + 4x + 5$, so it's $2\sqrt{x^2 + 4x + 5}$. Second piece: complete the square $(x + 2)^2 + 1$, giving $\ln|(x+2) + \sqrt{x^2 + 4x + 5}|$.

Putting it together: $2\sqrt{x^2 + 4x + 5} - \ln|(x+2) + \sqrt{x^2 + 4x + 5}| + C$.

Example 4. $\int \sqrt{9 - x^2}\,dx$.

By the standard form with $a = 3$: $\dfrac{x}{2}\sqrt{9 - x^2} + \dfrac{9}{2}\sin^{-1}(x/3) + C$.

Example 5. $\int \dfrac{dx}{x^2 - 6x + 13}$.

Complete the square: $(x - 3)^2 + 4$. So integral $= \dfrac{1}{2}\tan^{-1}\!\left(\dfrac{x - 3}{2}\right) + C$.

Example 6. $\int \dfrac{dx}{\sqrt{5x^2 - 2x}}$.

Factor inside: $5x^2 - 2x = 5\left(x^2 - \dfrac{2x}{5}\right) = 5\left[\left(x - \dfrac{1}{5}\right)^2 - \dfrac{1}{25}\right]$. So integrand is $\dfrac{1}{\sqrt{5}\sqrt{(x - 1/5)^2 - 1/25}}$, giving $\dfrac{1}{\sqrt{5}}\ln\left|(x - 1/5) + \sqrt{(x-1/5)^2 - 1/25}\right| + C$.

Try it yourself

1. $\int \dfrac{dx}{x^2 + 4x + 8}$
2. $\int \dfrac{dx}{\sqrt{16 - x^2}}$
3. $\int \dfrac{dx}{\sqrt{x^2 + 6x + 13}}$
4. $\int \dfrac{2x - 1}{x^2 + 2x + 5}\,dx$
5. $\int \dfrac{x + 1}{\sqrt{1 - 4x^2}}\,dx$
6. $\int \sqrt{x^2 + 2x + 5}\,dx$ (complete the square, then standard form)
7. $\int \dfrac{dx}{3x^2 - 5}$
8. $\int \sqrt{1 - x - x^2}\,dx$
9. $\int \dfrac{dx}{(x+1)\sqrt{x^2 + 2x + 1}}$ , note the singularity!
10. $\int \dfrac{3x + 2}{\sqrt{4x - x^2}}\,dx$
11. $\int \dfrac{x^2}{\sqrt{x^2 + 1}}\,dx$ (write $x^2 = (x^2 + 1) - 1$)
12. $\int \sqrt{4 - 9x^2}\,dx$
13. $\int \dfrac{dx}{(x - 1)\sqrt{x^2 - 1}}$ (substitute $x = \sec\theta$)
14. $\int \dfrac{2x + 5}{\sqrt{x^2 + 5x - 6}}\,dx$

Pitfalls and tricks

• Always complete the square first for quadratics in denominators.
• Pull out the leading coefficient before applying a standard form , $5x^2$ is not $x^2$.
• For linear-over-quadratic, split into a "derivative of denominator" piece and a constant piece.
• The radical forms with $\ln$ have no absolute-value confusion when $a > 0$ and the expression under the root is positive , but the natural domain restriction comes from the radicand.
• Memorise the table. These integrals appear in nearly every JEE paper.