# How to solve: $\int{\frac{r}{\sqrt{a r^2 + b r + c}}}dr$

I've been trying to solve the following integral: $$\int{\frac{r}{\sqrt{a r^2 + b r + c}}}dr$$ But I don´t get the substitution trick that I have to perform. I've tried to: $$r^2 = x$$ $$2r dr = dx$$ $$\frac{1}{2}\int{\frac{dx}{\sqrt{a x + b \sqrt{x} + c}}}$$ But I also do not know how to solve this. I've also tried to do decomposition in simple fractions, but this won´t work because of the root. I would apreciate any hint on what variable change I should perform.

• Euler substitutions rationalize all integral of that form and some.
– NDB
Commented Aug 12, 2023 at 15:42
• Do you know anything about the coefficients $a, b, c$, most importantly the values of the sign $\operatorname{sgn}(a)$ of $a$ and the discriminant $b^2 - 4 a c$? Those two values determine the character of the solution and, e.g., which trigonometric or Euler substitution is appropriate. Commented Aug 13, 2023 at 22:35

$$\newcommand{mbrack}[3]{\left#1 #2 \right #3} \newcommand{|}[1]{\mbrack{\lvert}{#1}{\rvert}}$$ first,factoring out the $$a$$ in the square root $$\int \frac{rdr}{\sqrt{ar^2+br+c}}=\frac{1}{\sqrt{a}}\int \frac{rdr}{\sqrt{r^2+\frac{b}{a}r+\frac{c}{a}}}$$ let $$I=\int \frac{rdr}{\sqrt{r^2+\frac{b}{a}r+\frac{c}{a}}}$$ since $$(a+b)^2=a^2+2ab+b^2$$ then $$\mbrack{(}{a+\frac{b}{2}}{)}^2=a^2+ab+\frac{b^2}{4}$$ completing the square $$r^2+\frac{b}{a}r+\frac{c}{a}=\mbrack{(}{r+\frac{b}{2a}}{)}^2+\frac{c}{a}-\frac{b^2}{4}$$ let \begin{align} s&=\frac{b}{2a}\\ q&=\frac{c}{a}-\frac{b^2}{4} \end{align} $$I=\int \frac{rdr}{\sqrt{(r+s)^2+q}}$$ substituting \begin{align} t&=r+s\\ r&=t-s\\ dt&=dr \end{align} $$I=\int \frac{tdt}{\sqrt{t^2+q}}-s\int \frac{dt}{\sqrt{t^2+q}}≝J_1-sJ_2$$ substituting again \begin{align} x&=t^2+q\\ dx&=2tdt\\ \frac{dx}{2}&=tdt \end{align} using the power rule and resubstituting $$J_1=\frac{1}{2}\int x^{-\frac{1}{2}}dx=\sqrt x=\sqrt{t^2+q}=\frac{1}{2}\sqrt{(r+s)^2+q}=\frac{1}{2}\sqrt{r^2+\frac{b}{a}r+\frac{c}{a}}$$ factoring out the $$\sqrt q$$ and substituting $$J_2=\sqrt q\int \frac{dt}{\sqrt{\mbrack{(}{\frac{t}{\sqrt{q}}}{)}^2+1}}$$ \begin{align} u&=\frac{t}{\sqrt{q}}\\ du&=\frac{1}{\sqrt{q}}dt\\ dt&=\sqrt q du \end{align} $$J_2=q\int \frac{1}{\sqrt{u^2+1}}du$$ using a trig substitution \begin{align} u&=\tan(\theta)\\ du&=\sec^2(\theta)d\theta \end{align} using the trig identity and multiplying top and bottom by $$\sec\theta + \tan\theta$$ $$J_2=q\int \frac{\sec^2 \theta d\theta}{\sqrt{1+\tan^2 \theta}}=q\int \sec\theta d\theta=q\int \frac{\sec\theta(\sec\theta+\tan\theta)d\theta}{\sec\theta+\tan\theta}$$ let \begin{align} w&=\sec\theta+\tan\theta\\ dw&=(\sec\theta\tan\theta+\sec^2\theta)d\theta=\sec\theta(\sec\theta+\tan\theta)d\theta \end{align} integrating and resubstituting $$J_2=q\int\frac{dw}{w}=q\ln\mbrack{\lvert}{w}{\rvert}=q\ln\|{\sec\theta+\tan\theta}=q\ln\|{\sec(\tan^{-1}(u))+u}$$ let \begin{align} \varphi&=\tan^{-1}(u)\\ \tan(\varphi)&=\frac{\alpha}{\beta}\\ \alpha&=u\\ \beta&=1\\ \sec(\varphi)&=\frac{\gamma}{\beta}=\alpha^2+\beta^2=u^2+1 \end{align} where $$\alpha,\beta,\gamma$$ are the opposite,adjacent and hypotenuse side of the right triangle with angle $$\varphi$$ $$\\$$ continuing to resubstitute $$J_2=q\ln\|{u^2+u+1}=q\ln\mbrack{\lvert}{\frac{t^2}{q}+\frac{t}{\sqrt q}+1}{\rvert}=q\ln\mbrack{\lvert}{\frac{1}{q}\mbrack{(}{t^2+\sqrt{q}t+q}{)}}{\rvert}\\=q\ln\mbrack{\lvert}{\frac{1}{q}\mbrack{(}{r^2+\mbrack{(}{\sqrt{q}+\frac{b}{a}}{)}r+s+q}{)}}{\rvert}=\mbrack{(}{\frac ca-\frac{b^2}{4}}{)}\ln\mbrack{\lvert}{r^2+\mbrack{(}{\sqrt{\frac{c}{a}-\frac{b^2}{4}}+\frac{b}{a}}{)}r+\frac{c}{a}-\frac{b^2}{4}+\frac{b}{2a}}{\rvert}$$ $$I=\frac{1}{2}\sqrt{r^2+\frac{b}{a}r+\frac{c}{a}}-\frac{b}{2a}\mbrack{(}{\frac ca-\frac{b^2}{4}}{)}\ln\mbrack{\lvert}{r^2+\mbrack{(}{\sqrt{\frac{c}{a}-\frac{b^2}{4}}+\frac{b}{a}}{)}r+\frac{c}{a}-\frac{b^2}{4}+\frac{b}{2a}}{\rvert}+K,a\ne0$$ $$\int\frac{rdr}{\sqrt{br+c}}=\frac{1}{\sqrt b}\int\frac{rdr}{\sqrt{r+\frac{c}{b}}}$$ (where $$K$$ is an arbitrary constant) note that because there are divisions by $$a$$ ,therefore the above formula only works for $$a\ne0$$ $$\\$$ now doing the case for $$a=0$$ $$\xi=\int\frac{rdr}{\sqrt{r+\frac{c}{b}}}$$ \begin{align} \tau&=r+\frac{c}{b}\\ d\tau&=dr \end{align} using the power rule $$\xi=\int \frac{\tau d\tau}{\sqrt \tau}+\frac{c}{b}\int\frac{d\tau}{\sqrt \tau}=\frac{2}{3}\tau^{3/2}+\frac{2c}{b}\sqrt{\tau}=\frac{2}{3}\mbrack{(}{r+\frac{c}{b}}{)}^{3/2}+\frac{2c}{b}\sqrt{r+\frac{c}{b}}+K,b\ne0$$ again, since there are divisions by $$b$$ the above formula only works for $$b\ne0$$ $$\\$$ doing the case for $$b=0$$ $$\int \frac{rdr}{\sqrt c}=\frac{1}{\sqrt c}\int rdr=\frac{1}{2c}r^2+K$$ finally $$\int \frac{rdr}{\sqrt{ar^2+br+c}}=\begin{cases}\frac{1}{2\sqrt a}\sqrt{r^2+\frac{b}{a}r+\frac{c}{a}}-\frac{b}{2a}\mbrack{(}{\frac ca-\frac{b^2}{4}}{)}\ln\mbrack{\lvert}{r^2+\mbrack{(}{\sqrt{\frac{c}{a}-\frac{b^2}{4}}+\frac{b}{a}}{)}r+\frac{c}{a}-\frac{b^2}{4}+\frac{b}{2a}}{\rvert}+K\hspace{15mm}&a\ne0\\\frac{2}{3\sqrt b}\mbrack{(}{r+\frac{c}{b}}{)}^{3/2}+\frac{2c}{b\sqrt b}\sqrt{r+\frac{c}{b}}+K &a=0,b\ne0\\\frac{1}{2c}r^2+K &a=0,b=0\end{cases}$$
Assuming $$a\ne 0$$ $$\int{\frac{r}{\sqrt{a r^2 + b r + c}}}dr = \frac1a \sqrt{a r^2 + b r + c} -\frac b{2a}\int \frac1 {\sqrt{a r^2 + b r + c}}dr$$ Then, substitute $${\sqrt{a r^2 + b r + c}}=(ar+\frac b2)t$$ \begin{align} \int \frac1 {\sqrt{a r^2 + b r + c}}dr =&\int\frac1{1-at^2}dt\\ =&\begin{cases} \frac1{\sqrt{-a}}\tan^{-1}\frac{\sqrt{-a} \sqrt{a r^2 + b r + c} }{ar+\frac b2},&a<0 \\ \frac1{2\sqrt a}\ln\bigg|\frac{ar+\frac b2 +\sqrt{a} \sqrt{a r^2 + b r + c} }{ar +\frac b2-\sqrt{a} \sqrt{a r^2 + b r + c} }\bigg|,&a>0 \end{cases} \end{align}
Hint: Try to write the integral in the form $$\int \frac{f'(x)dx}{ \sqrt{f(x)}}.$$ Here, your question, think which thing you have to multiply, divide , add and subtract to get the above form.