Find the indefinite integral of (1 - u^2) divided by (a - b u^2) raised to the 3/2 power, with respect to u.

Step-by-step solution

1. 1

   Build the key antiderivative by differentiating a guess. Rather than a trig substitution, differentiate $\dfrac{u}{\sqrt{a-bu^2}}$ using the product rule:

   $\frac{d}{du}\left[u(a-bu^2)^{-1/2}\right]=\frac{1}{\sqrt{a-bu^2}}+\frac{bu^2}{(a-bu^2)^{3/2}}=\frac{a}{(a-bu^2)^{3/2}}$

   Hence

   $\int\frac{du}{(a-bu^2)^{3/2}}=\frac{u}{a\sqrt{a-bu^2}}$

2. 2

   Split the numerator.

   $\int\frac{1-u^2}{(a-bu^2)^{3/2}}\,du=\int\frac{du}{(a-bu^2)^{3/2}}-\int\frac{u^2\,du}{(a-bu^2)^{3/2}}$

   The first piece is already done; only the $u^2$ piece needs work.

3. 3

   Turn $u^2$ into the denominator's own expression. The identity

   $u^2=\frac{a-(a-bu^2)}{b}$

   lets us write

   $\frac{u^2}{(a-bu^2)^{3/2}}=\frac{a}{b(a-bu^2)^{3/2}}-\frac{1}{b\sqrt{a-bu^2}}$

   This is the whole trick: it converts an awkward numerator into two integrals we already recognise, avoiding any trigonometric substitution.

4. 4

   Assemble the pieces. Using the Step 1 result and $\displaystyle\int\frac{du}{\sqrt{a-bu^2}}=\frac{1}{\sqrt b}\arcsin\!\left(u\sqrt{\tfrac{b}{a}}\right)$:

   $I=\frac{u}{a\sqrt{a-bu^2}}-\frac{u}{b\sqrt{a-bu^2}}+\frac{1}{b\sqrt b}\arcsin\!\left(u\sqrt{\tfrac{b}{a}}\right)+C$

   Combining the first two terms over $ab$ gives the coefficient $\dfrac{b-a}{ab}$.

5. 5

   Verify by differentiating the answer. Take the concrete values $a=2$, $b=0.7$. Numerically differentiating the closed form at $u=0.1$ gives $0.3518635229$, and the integrand $(1-u^2)/(a-bu^2)^{3/2}$ at $u=0.1$ is $0.3518635229$ ✓. At $u=0.5$ both give $0.3042054288$ ✓. The antiderivative is correct.