What is ∫{dx}{{x²+a²}} equal to?

Correct Answer is : ln∣x+x2+a2a∣+cln | {x+{x²+a²}}{a} | +cln∣ax+x2+a2∣+c

Correct Answer is : ln∣x+x2+a2a∣+cln | {x+{x²+a²}}{a} | +cln∣ax+x2+a2∣+c

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UPSC NDA I 2015 Mathematics Paper

Question : 86 of 120

Marks: +1, -0

What is

∫\frac{dx}{\sqrt{x^{2}+a^{2}}}

equal to?

Solution:

Concept:

Use the substitution

x = a \tan \theta

to simplify the integral.

Explanation:

Let

x = a \tan \theta

, so

dx = a \sec^2 \theta \, d\theta

Then

\sqrt{x^2 + a^2} = \sqrt{a^2 \tan^2 \theta + a^2} = a \sec \theta

The integral becomes:

\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a \sec^2 \theta \, d\theta}{a \sec \theta} = \int \sec \theta \, d\theta

We know

\int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| + C

Back-substitute:

\tan \theta = \frac{x}{a}

\sec \theta = \sqrt{1 + \tan^2 \theta} = \frac{\sqrt{x^2 + a^2}}{a}

Thus

\sec \theta + \tan \theta = \frac{\sqrt{x^2 + a^2} + x}{a}

Therefore,

\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|\frac{x + \sqrt{x^2 + a^2}}{a}\right| + C

Answer:

Option A:

\ln\left|\frac{x + \sqrt{x^2 + a^2}}{a}\right| + C

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