Question

$\int \frac{dx}{x^4\sqrt{a^2+x^2}}=\frac{1}{a^4}[\frac{1}{x}\sqrt{a^2+x^2}-\frac{1}{3x^3}{(a^2+x^2)}^{3/2}]$. If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \frac{dx}{x^4\sqrt{a^2+x^2}}$

$=\int \frac{dx}{x^5\sqrt{(\frac{a^2}{x^2}+1)}}$

$=\int \frac{dx}{x^3.x^2\sqrt{\frac{a^2}{x^2}+1}}$
Substitute $\frac{a^2}{x^2}+1=t^2\Rightarrow -\frac{2a^2}{x^3}dx=2tdt$
$\therefore I=\int -\frac{1}{a^2}t.\frac{(t^2-1)}{a^2}.\frac{1}{t}dt=-\frac{1}{a^4}(\frac{t^3}{3}-t)=\frac{1}{a^4}[t-\frac{t^3}{3}]$
Where $t=\sqrt{\frac{a^2}{x^2}+1}=\frac{\sqrt{a^2+x^2}}{x}$

$\therefore$$\int \frac{dx}{x^4\sqrt{a^2+x^2}}=\frac{1}{a^4}[\frac{1}{x}\sqrt{a^2+x^2}-\frac{1}{3x^3}{(a^2+x^2)}^{3/2}]$