Question

Solve it by differentiating on Right hand side rather than integrating on the left hand side.

Answer 1

Solution

Take LHS

$\to\bf \int\dfrac{dx}{x\sqrt{1-x^3} }$

Now Multiply and divide  by x²

$\to\bf \int\dfrac{x^2dx}{x^3\sqrt{1-x^3} }$

Now using Substitution method

$\to\bf\sqrt{1-x^3} =t$

$\bf\to \dfrac{d(\sqrt{1-x^3} )}{dx} =\dfrac{dt}{dx}$

$\to\bf \dfrac{-3x^2}{2\sqrt{1-x^2} } dx=dt$

$\to\bf x^2dx=\dfrac{-2}{3} \sqrt{1-x^3} dt$

We can Write as $\bf\sqrt{1-x^3} =t$

$\to\bf x^2dx=\dfrac{-2}{3} tdt$

To Eliminate x³ we can write as

$\to\bf\sqrt{1-x^3} =t$

$\to\bf{1-x^3} =t^2$

$\to\bf{x^3} =1-t^2$

Now

$\to\bf \int\dfrac{x^2dx}{x^3\sqrt{1-x^3} }$

Now by putting the value we get

$\to\bf\int\dfrac{-2tdt}{3(1-t^2)t} =\dfrac{-2}{3} \int\dfrac{dt}{(1-t^2)}$

Now take common ' - ' from denominator , we get

$\to\bf\dfrac{2}{3} \int\dfrac{dt}{(t^2-1)}$

Now using the identities (t²-1²)=(t-1)(t+1)

$\to\bf\dfrac{2}{3} \int\dfrac{dt}{(t-1)(t+1)}=\dfrac{1}{3} \int\dfrac{(t-1)-(t+1)}{(t-1)(t+1)}dt$

$\bf\to\dfrac{1}{3} \bigg\{\int\dfrac{(t-1)}{(t-1)(t+1)}dt-\int\dfrac{(t+1)}{(t-1)(t+1)} dt\bigg\}$

$\bf\to\dfrac{1}{3} \bigg\{ \int\dfrac{dt}{(t-1)} -\int\dfrac{dt}{(t+1)} \bigg\}$

$\bf\to\dfrac{1}{3} \bigg(ln|t-1|-ln|t+1|\bigg)+c$

By Using the log properties we can write as

$\bf\to\dfrac{1}{3} ln\bigg|\dfrac{t-1}{t+1} \bigg|+c$

Now put the value of t , we get

$\bf\to\dfrac{1}{3} ln\bigg|\dfrac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} +1} \bigg|+c$

Answer

So value of a = 1/3  and  b = -1