Integrals of Some Particular Functions

A standard integral is an integral whose result is already known and can be directly used after suitable algebraic simplification or substitution. In this topic, the main idea is to convert complicated expressions into one of the known standard forms.

• \[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C\]
• \[\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C\]
• \[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\]
• \[\int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C\]
• \[\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C\]
• \[\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C\]

• Partial Fractions: Used for standard rational functions like \[\frac{1}{x^2-a^2}\] by splitting the denominator into linear components: \[(x-a)(x+a)\].

• Trigonometric Substitution: Used for expressions containing radical terms or squares:

  • For \[\sqrt{a^2 - x^2}\] or \[a^2 - x^2\], substitute $x = a \sin \theta$.

  • For \[x^2 + a^2\] or \[\sqrt{x^2 + a^2}\], substitute \[x = a \tan \theta\].

  • For \[x^2 - a^2\] or \[\sqrt{x^2 - a^2}\], substitute \[x = a \sec \theta\].

When integrals appear in the form \[\int \frac{dx}{ax^2 + bx + c}\] or \[\int \frac{dx}{\sqrt{ax^2 + bx + c}}\], use the completing the square method to transform the quadratic expression into \[(x \pm h)^2 \pm k^2\], making it compatible with the standard formulae above.

Find the following integrals

1. \[\int \frac{dx}{x^2 - 6x + 13}\]
2. \[\int \frac{dx}{3x^2 + 13x - 10}\]

Solution:

(i) We have \[x^2 - 6x + 13 = x^2 - 6x + 3^2 - 3^2 + 13 = (x - 3)^2 + 4\]

So, \[\int \frac{dx}{x^2 - 6x + 13} = \int \frac{1}{(x - 3)^2 + 2^2} dx\]

Let \[x - 3 = t\]. Then \[dx = dt\]

Therefore, \[\int \frac{dx}{x^2 - 6x + 13} = \int \frac{dt}{t^2 + 2^2} = \frac{1}{2} \tan^{-1} \frac{t}{2} + \text{C}\] [by 7.4 (3)]

\[= \frac{1}{2} \tan^{-1} \frac{x - 3}{2} + \text{C}\]

(ii) We write the denominator of the integrand,

\[3x^2 + 13x - 10 = 3 \left( x^2 + \frac{13x}{3} - \frac{10}{3} \right)\]

\[= 3 \left[ \left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2 \right]\] (completing the square)

Thus \[\int \frac{dx}{3x^2 + 13x - 10} = \frac{1}{3} \int \frac{dx}{\left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2}\]

Put \[x + \frac{13}{6} = t\] . Then \[dx = dt\].

Therefore, \[\int \frac{dx}{3x^2 + 13x - 10} = \frac{1}{3} \int \frac{dt}{t^2 - \left( \frac{17}{6} \right)^2}\]

\[= \frac{1}{3 \times 2 \times \frac{17}{6}} \log \left| \frac{t - \frac{17}{6}}{t + \frac{17}{6}} \right| + \text{C}_1\] [by 7.4 (i)]

\[= \frac{1}{17} \log \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right| + \text{C}_1\]

\[= \frac{1}{17} \log \left| \frac{6x - 4}{6x + 30} \right| + \text{C}_1\]

\[= \frac{1}{17} \log \left| \frac{3x - 2}{x + 5} \right| + \text{C}_1 + \frac{1}{17} \log \frac{1}{3}\]

\[= \frac{1}{17} \log \left| \frac{3x - 2}{x + 5} \right| + \text{C}\] , where \[\text{C} = \text{C}_1 + \frac{1}{17} \log \frac{1}{3}\]

Find the following integrals:

(i) \[\int \frac{x + 2}{2x^2 + 6x + 5} dx\] 

Solution:

(i) Using the formula 7.4 (9), we express

\[x + 2 = \text{A} \frac{d}{dx} (2x^2 + 6x + 5) + \text{B} = \text{A} (4x + 6) + \text{B}\]

Equating the coefficients of \[x\] and the constant terms from both sides, we get

\[4\text{A} = 1\] and \[6\text{A} + \text{B} = 2\] or \[\text{A} = \frac{1}{4}\] and \[\text{B} = \frac{1}{2}\].

Therefore, \[\int \frac{x + 2}{2x^2 + 6x + 5} = \frac{1}{4} \int \frac{4x + 6}{2x^2 + 6x + 5} dx + \frac{1}{2} \int \frac{dx}{2x^2 + 6x + 5}\]

\[= \frac{1}{4} \text{I}_1 + \frac{1}{2} \text{I}_2\] {0.5cm} (say)... (1)

In \[\text{I}_1\], put \[2x^2 + 6x + 5 = t\], so that \[(4x + 6) dx = dt\]

Therefore, \[\text{I}_1 = \int \frac{dt}{t} = \log |t| + \text{C}_1\]

\[= \log |2x^2 + 6x + 5| + \text{C}_1\]  ... (2)

and \[\text{I}_2 = \int \frac{dx}{2x^2 + 6x + 5} = \frac{1}{2} \int \frac{dx}{x^2 + 3x + \frac{5}{2}}\]

\[= \frac{1}{2} \int \frac{dx}{\left( x + \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2}\]

Put \[x + \frac{3}{2} = t\], so that \[dx = dt\], we get

\[\text{I}_2 = \frac{1}{2} \int \frac{dt}{t^2 + \left( \frac{1}{2} \right)^2} = \frac{1}{2 \times \frac{1}{2}} \tan^{-1} 2t + \text{C}_2\] [by 7.4 (3)]

\[= \tan^{-1} 2 \left( x + \frac{3}{2} \right) + \text{C}_2 = \tan^{-1} (2x + 3) + \text{C}_2\] ... (3)

Using (2) and (3) in (1), we get

\[\int \frac{x + 2}{2x^2 + 6x + 5} dx = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1} (2x + 3) + \text{C}\]

where, \[\text{C} = \frac{\text{C}_1}{4} + \frac{\text{C}_2}{2}\]

• Convert the integrand into a known standard form before integrating.

• For \[x^2 - a^2\], factorize and use partial fractions.

• For \[x^2 + a^2\], the answer usually involves \[\tan^{-1}\].

• For \[\sqrt{a^2 - x^2}\], the answer usually involves \[\sin^{-1}\].

• For general quadratics, complete the square first.

• For \[px + q\] in the numerator, relate it to the derivative of the denominator.

• Always write the constant of integration C in the final answer.