# indefinite integral $\int\sqrt{a+bx^2}dx$, a and b are constant arbitrary real numbers [closed]

Is there a general form for the indefinite integral:

$\int \sqrt{a+bx^2}$ dx

where a and b are constant arbitrary real numbers?

• These are often done with trigonometric substitution. The substitution depends on whether $a$ and $b$ are positive or negative. Jan 28 '18 at 21:56
• @Matthew Leingang thanks, can u give the cases, and I'll try the trig substitution on my own? Jan 28 '18 at 21:58
• Just google "trigonometric substitution"—any of the links on the first page are relevant. Jan 28 '18 at 22:01

Some common cases are: 1) if the inside of the integral is sqrt( a^2 - x^2) you can use x=asinu 2)if the inside of the integral is sqrt( a^2 +x^2) you can use x=atanu 3)if the inside of the integral is sqrt( x^2-a^2) you can use x=a/cosu

If $a,b>0$ it's the same integral as $\int\sqrt{1+x^2}$, if a,b have different sign then it's the same integral as $\int \sqrt{1-x^2}$ exept for a constant that is multiplied to the primitive.

1) After substitution, try with x=tan(u)

2) Try x=cos(u)

Hint: Use $$x=\sqrt{\dfrac{a}{b}}\tan u$$when $a,b>0$. Use $$\sqrt{\dfrac{a}{b}}\sin u$$when $a>0,b<0$ and $x=\sqrt{\dfrac{a}{b}}\sec u$ when $a<0,b>0$.

$\int \sqrt{a+bx^2}=\frac{x\sqrt{a+bx^2}}{2}+\frac{aln(\sqrt{b}x+\sqrt{a+bx^2})}{2\sqrt b}$

First substitute $$x=\sqrt{\frac{a}{b}}\sinh(u)\qquad dx=\sqrt{\frac{a}{b}}\cosh(u)$$ Your integral becomes \begin{align*}\\ I&=\int\sqrt{a+bx^2}\,dx\\ I&=\frac{a\sqrt{b}}{b}\int \sinh^2 (u)\,du\\ I&=\frac{a\sqrt{b}}{2b}\int \cosh(2u)-1\,du\\ I&=\frac{a\sqrt{b}}{2b}\left(\frac{1}{2}\sinh(2u)-u\right)+C\\ I&=\frac{a\sqrt{b}}{2b}\left( \frac{1}{2}\sinh\left(2\sinh^{-1}\left(\sqrt{\frac{b}{a}} x\right)\right)-\sinh^{-1}\left(\sqrt{\frac{b}{a}}x\right)\right)+C \end{align*}