Question

How many of the following integrals are correct?
1. $\int \frac{dx}{\sqrt{x^2+a^2}}=ln|x+\sqrt{x^2+a^2}|+C$
2. $\int \frac{dx}{\sqrt{x^2-a^2}}=ln|x-\sqrt{x^2-a^2}|+C$
3. $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}ln\left|\frac{x-a}{x+a}\right|+C$
4. $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}ln\left|\frac{x+a}{x-a}\right|+C$
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Answer 1

All the above integrals are standard integrals . We will now try to derive these formulae using substitution. But it is recommended to remember these formulae.
1. $\int \frac{dx}{\sqrt{x^2+a^2}}$
Since we have the term, $x^2$ + $a^2$, we will go for the substitution $x=atan\theta$. We can also go for the substitution $x=acot\theta$. We do this because $x^2$ + $a^2$ gives a familiar trigonometric function as we proceed.
$x=atan\theta$
$\Rightarrow dx=ase{c}^2\theta .d\theta$
Also,
$x^2+a^2=(atan\theta {)}^2+a^2=a^2(tan{\theta }^2+1)=a^{2}se{c}^2\theta \Rightarrow \int \frac{dx}{\sqrt{x^2+a^2}}=\int \frac{ase{c}^2\theta .d\theta }{\sqrt{a^{2}se{c}^2\theta }}=\int |sec\theta |d\theta =ln|sec\theta +tan\theta |+c$
We have $tan\theta =\frac{x}{a}$ from the substitution we made. We can find $sec\theta$ using the relation $sec{\theta }^2=tan{\theta }^2+$
1.
So, we get $sec\theta =\frac{\sqrt{x^2+a^2}}{a}$.
Substituting these values we get
$\int \frac{dx}{\sqrt{x^2+a^2}}=ln|sec\theta +tan\theta |=ln\left|\frac{x+\sqrt{x^2+a^2}}{a}\right|+c=ln|x+\sqrt{x^2+a^2}|-ln|a|+c$
We can combine ‘ - ln |a|’ and ‘c’ and call it as C, because both are constants. So, the first integral is correct.
2. We can find the integral $\int \frac{dx}{\sqrt{x^2-a^2}}$ in a similar way with the substitution,
$x=asec\theta$
We will get,
$\int \frac{dx}{\sqrt{x^2-a^2}}=ln|x+\sqrt{x^2-a^2}|+C$
This is different from the integral given in statement 2. So second statement is wrong.
3. $\int \frac{dx}{x^2-a^2}$
We will solve this question by rearranging the terms.
We can write,
$\frac{1}{x^2-a^2}=\frac{1}{2a}[\frac{1}{x-a}-\frac{1}{x+a}]$
So we get,
$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\left[\int \frac{1}{x-a}-\int \frac{1}{x+a}\right]=\frac{1}{2a}\left[ln(|x-a|-ln|x+a|)\right]+C=\frac{1}{2a}\left[ln\left|\frac{x-a}{x+a}\right|\right]+C$
Rearranging terms in this way and finding integral is covered in integration by partial fractions in detail
4. We can find this integral in the same way we found the third integral. We get $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}ln\left|\frac{x+a}{x-a}\right|+C$