$$\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx$$

So this is what I did:

$u = 3ax + bx^3$

$\dfrac{\mathrm du}{\mathrm dx}= 3a + 3bx^2$

$\mathrm du = 3a + 3bx^2 dx$

= $\displaystyle\int 3\cdot \frac{1}{\sqrt{u}}\mathrm du$

= $3\cdot \frac{u^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c$

= $3\cdot \frac{\left(3ax+bx^3\right)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c$

but this is incorrect.

The right answer is actually


which is not equivalent to the above answer even in simplest form.

Any help?

  • $\begingroup$ In fact, they are exactly the same answer. $\endgroup$ – Zamu Jan 25 '18 at 11:08

$$\begin{align}I&=\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx\\&=\dfrac13\int\dfrac{3a+3bx^2}{\sqrt{3ax+bx^3}}\,\mathrm dx\\&=\dfrac13\int\dfrac{\mathrm d(3ax+bx^3)}{\sqrt{3ax+bx^3}}\qquad\text{Use substitution now, let }u=3ax+bx^3\\&=\dfrac13\int\dfrac{\mathrm du}{\sqrt u}\\&=\dfrac13\cdot 2\sqrt u+C\\&=\dfrac23\sqrt{3ax+bx^3}+C\end{align}$$


Note that $$du = (3a + 3bx^2) dx \implies dx=(1/3)(a + bx^2)du $$

That is how you get your $(1/3)$ in the answer.


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