Indefinite integral question: $\int \frac{1}{x\sqrt{x^2+x}}dx$

Question

How can I solve this integral: $$\int \frac{1}{x\sqrt{x^2+x}}dx$$ I first completed the square and got: $$\int \frac{1}{x\sqrt{(x+\frac{1}{2})^2-\frac{1}{4}}}dx$$ Then I factored out 1/4 and got: $$2\int \frac{1}{x\sqrt{(2x+1)^2-1}}dx$$ Then I substituted $2x+1$ with $t$ and got: $$2\int \frac{1}{(t-1)\sqrt{t^2-1}}dt$$ I'm not sure what to do next. Please give me a hint ;)

Answer 1

Here is an approach.

$$ \begin{align} \int \frac{1}{x \sqrt{x^2+x}}dx &=\int \frac{1}{\sqrt{1+\frac{1}{x}}}\frac{dx}{x^2}\\\\ &=-\int \frac{1}{\sqrt{1+u}}\:du \quad (u=1/x)\\\\ &=-2\sqrt{u+1}+C\\\\ &=-2\sqrt{\frac{x+1}{x}}+C \end{align} $$ where $C$ is any constant.

Answer 2

If you don't want to have to "spot" which substitution to use, but instead want a "cookbook" approach, then consider using an Euler substitution. This would work here because you have a rational function of $x$ and $\sqrt{ax^2 + bx + c}$, in particular with $a=1$, $b=1$, and $c=0$.

You can use Euler's first substitution if $a>0$: here we write $\sqrt{ax^2+bx+c} = \pm x\sqrt{a}+t$ which gives $x = \frac{c-t^2}{\pm 2t\sqrt{a}-b}$. Since $a=1$ that's an option, but neither an $x+t$ nor $-x+t$ in the denominator look particularly helpful here because of the multiplication by $x$. If that had been an addition or subtraction of $x$ instead, then this substitution would have been ideal as that would have cancelled.

According to Wikipedia, we can use Euler's second substitution if $c>0$: here we write $\sqrt{ax^2+bx+c} = xt\pm\sqrt{c}$ which gives $x = \frac{\pm 2t\sqrt{c}-b}{a-t^2}$. Since $c=0$ I will not pursue this.

Euler's third substitution is used if $ax^2+bx+c$ has real roots $\alpha$ and $\beta$; we write $\sqrt{ax^2 + bx + c} = \sqrt{a(x-\alpha)(x-\beta)} = (x-\alpha)t$ which gives $x = \frac{a\beta-\alpha t^2}{a-t^2}$. This multiplicative form seems more suited for our purposes. In our case $x^2 + x = x(x+1)$ so we can take $\alpha=0$ and $\beta=-1$.

Pursuing the third substitution we obtain immediately $\sqrt{x^2 + x} = \sqrt{x(x+1)} = xt$ whence $t=\frac{\sqrt{x(x+1)}}{x} = \sqrt{\frac{x+1}{x}}$ and $x=\frac{1(-1) - 0t^2}{1-t^2}=(t^2-1)^{-1}$. Then $\frac{\mathrm{d}x}{\mathrm{d}t}=-2t(t^2-1)^{-2}$. Putting this together,

$$\int \frac{\mathrm{d}x}{x\sqrt{x(x+1)}}= \int \frac{-2t(t^2-1)^{-2}\mathrm{d}t}{x^2t} = \int \frac{-2t(t^2-1)^{-2}\mathrm{d}t}{(t^2-1)^{-2} t} = \int -2 \mathrm{d}t = -2t + C$$

Expressing back in terms of $x$ gives $-2 \sqrt{\frac{x+1}{x}} + C$. The advantage of this approach is that no "cleverness" is needed to find the required substitution, we just work through a checklist and see which one applies. A cleverer solution might be faster to perform, but take longer to spot.

Going back to Euler's second substitution, we can see that even though $c=0$ this would have worked just fine, and in fact it would have immediately given the same result as Euler's third substitution. I have the feeling the "$>$" in the Wikipedia article should really say "$\geq$" - but it doesn't make difference, since when $c=0$ the quadratic would factorise with one of the roots as zero and we can apply the third substitution as above.

Finally, reconsidering the first substitution, we could have gone ahead and put $\sqrt{x^2 + x} = x + t$ and $x = \frac{-t^2}{2t-1} = t^2 (1-2t)^{-1}$. By product rule,

$$\frac{\mathrm{d}x}{\mathrm{d}t}=2t(1-2t)^{-1} + 2t^2(1-2t)^{-2} = 2t(1-t)(1-2t)^{-2}$$

Substituting into the integral and clearing up is rather messy, for the reasons mentioned above, but can be done. Skipping quite a lot of algebra,

$$\int \frac{\mathrm{d}x}{x\sqrt{x(x+1)}}= \int \frac{2t(1-t)(1-2t)^{-2}\mathrm{d}t}{t^2 (1-2t)^{-1} ( t^2 (1-2t)^{-1} + t)} = \int 2t^{-2} \mathrm{d}t = -2t^{-1} + C$$

In terms of $x$, this is $\frac{2}{x - \sqrt{x^2 + x}} + C$. This is an alternative form to that given by Euler's third substitution, but was harder work to acquire. So don't just jump straight for the first substitution you see that fulfils the necessary conditions on the coefficients; spend a moment considering which of the available options best fits the form of the question.

Answer 3

Apply the Euler substitution $\sqrt{x^2+x}=xt$ $$\int \frac{1}{x\sqrt{x^2+x}}dx =-2\int dt=-\frac{2 \sqrt{x^2+x}}x+C $$ which is valid for all domain $x$.

Answer 4

To do integrations of the form $$\int \frac {\mathrm dx}{P(x)\sqrt{ Q(x)}}$$ where $P(x)$ is linear and $Q(x)$ is quadratic, you can substitute $ P(x) = u $. Substituting $t-1 = \frac{1}{u}$ reduces the integral to $$\frac{1}{2}\int \frac {-\mathrm du}{\sqrt{1 +2u}}.$$ Now this can be integrated further.

Answer 5

This integral can be easily solved using exponential substitution (es). Applying the general transformation formula $(24)$ in this blog post, this is

$$\int f\left(x, \sqrt{(x+b)^2-a^2}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a-b, \frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2}a\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha,$$

where $a=\frac12$ and $b=\frac12$, the integral becomes

$$\begin{aligned}\int \frac{1}{x\sqrt{x^2+x}}\,dx&=\int\frac{1}{x\sqrt{\left(x+\frac12\right)^2-\frac14}}\,dx\\&\overset{es}{=}-4i\int\frac{1}{e^{i\alpha}+e^{-i\alpha}-2}\,d\alpha\\&=-4i\int\frac{e^{i\alpha}}{(e^{i\alpha}-1)^2}\,d\alpha\\&=\frac{4}{e^{i\alpha}-1}+C\end{aligned}$$ To switch to real numbers, plug in the values of $a$ and $b$ into formula $(22)$: $$e^{\text{i}\alpha}=\frac{x + \frac12 - \sqrt{x^2+x}}{\frac12}=2x+1-2\sqrt{x^2+x}$$ Thus, we obtain $$\int \frac{1}{x\sqrt{x^2+x}}\,dx=\frac{2}{x-\sqrt{x^2+x}}+C$$

---

From $(22)$, by introducing the substitution

$$t=e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b \mp \sqrt{(x + b)^2 - a^2}}{a},\tag{*}$$

we can reformulate the transformation formula $(24)$ so that extended Weierstrass substitution can be applied directly to integrals with irrational integrands. This approach is neater than employing complex exponentials. However, on the interval $[-1, 1]$, using complex exponentials may be more convenient, since the tangent function becomes undefined at $\frac{x+b}{a}=0$.

With this substitution, we have:

$$\int f\left( x, \sqrt{(x + b)^2 - a^2}, \dfrac{ \sqrt{x + b - a} }{ \sqrt{x + b + a} } \right) \, dx= \int f\left(\frac{t^2+1}{2t}a - b, \mp\frac{t^2-1}{2t}a, \pm\frac{1-t}{1+t}\right) \frac{t^2-1}{2 t^2}a\, dt.\tag{**}$$

For the alternating signs $\mp$ and $\pm$:

• Use the upper sign when $\dfrac{x + b}{a} \geq 1$

• Use the lower sign when $\dfrac{x + b}{a} < -1$

Similarly, transformation formula $(20)$ can be rewritten as follows

$$\int f\left( x, \sqrt{(x + b)^2 + a^2}\right) \, dx= \int f\left(\frac{s^2-1}{2s}a - b, \frac{s^2+1}{2s}a\right) \frac{s^2+1}{2 s^2}a\, ds,\tag{***}$$

where $s=\coth{\left(\frac12\text{csch}^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{x+b+\sqrt{(x+b)^2+a^2}}{a}$, for $\frac{x+b}{a}>0$ and $a>0$.

That said, for $x>0$, applying $(**)$ the OP's integral reduces to

$$\begin{aligned}\int \frac{1}{x\sqrt{x^2+x}}\,dx&=\int\frac{1}{x\sqrt{\left(x+\frac12\right)^2-\frac14}}\,dx\\&=\int \frac{1}{\bigl(\tfrac{t^2 + 1}{4t} - \tfrac{1}{2}\bigr)\,\bigl(\tfrac{1 - t^2}{4t}\bigr)} \cdot \frac{t^2 - 1}{4t^2}\,dt\\&=-4\int\frac{1}{(t-1)^2}\,dt\end{aligned},$$

which is easy to integrate. To express back in terms of $x$ you can use $(*)$ as in my previous solution. For $x<-1$, just use lower sign.

Interestingly, I’ve found that this extended Weierstrass substitution, for some cases, simplifies integrals into rational forms nearly identical to those obtained through Euler’s substitutions. Although I find the case distinction in this approach more intuitive, it also allows me to switch to complex exponentials or trigonometric/hyperbolic functions whenever I find it convenient. What I see here is an approach that unifies and extends trigonometric/hyperbolic substitutions, Weierstrass substitution, the use of sine and cosine as complex exponentials, and Euler’s substitutions (?) to irrational integrands. I’m in shock!

Answer 6

For $x>0,$ the substitution $x=\sinh^2u$ works nicely, $$\int\frac{dx}{x\sqrt{x^2+x}}= \int\frac{dx}{x^{\frac32}\sqrt{x+1}}\stackrel{x=\sinh^2u}{=}\int2\operatorname{csch}^2u\,du =-2\coth u+C=-2\frac{\sqrt{x+1}}{\sqrt{x}}+C $$

Answer 7

Take $x^2$ out from that under the square root sign and it will transform into $$ \int \frac{\mathrm dx}{x^2 \sqrt{1+\frac{1}{x}}}.$$ Now take $-$ common and take $1+1/x$ as $t^2$, then it becomes a trivial tast to perform the integration.

Answer 8

For yet another alternative, substitute $x=\dfrac{1-y}{1+y}$ :

$$\begin{align*} & \int \frac{dx}{x\sqrt{x^2+x}} \\ &= -2 \int \frac{1}{\frac{1-y}{1+y} \sqrt{\frac{(1-y)^2}{(1+y)^2}+\frac{1-y}{1+y}}} \, \frac{dy}{(1+y)^2} \\ &= -\sqrt2 \int \frac{dy}{(1-y)^{\frac32}} \\ &= -\frac{2\sqrt2}{\sqrt{1-y}} + C \\ &= -\frac{2\sqrt2}{\sqrt{1-\frac{1-x}{1+x}}} + C \\ &= -2\sqrt{\frac{1+x}x} + C \end{align*}$$

Answer 9

Well $9$ years is too old for me. But still, I want to come up with a new process for this integral.

Let's assume that $$I=\int\frac{1}{x\sqrt{x^{2}+x}}\mathrm dx$$

$$\implies I=\int\frac{1}{x\sqrt{x(x+1)}}\mathrm dx$$

$$\implies I=\int\frac{1}{x\sqrt{x}\sqrt{x+1}}\mathrm dx$$

$$\implies I=\int\frac{1}{\sqrt{x^{3}}\sqrt{x+1}}\mathrm dx$$

$$\implies I=\int\frac{1}{x^{\frac{3}{2}}(x+1)^{\frac{1}{2}}}\mathrm dx$$

Now, substitute $x=\tan^{2}(\theta)$ $\implies\mathrm dx=2\tan(\theta)\sec^{2}(\theta)\mathrm d\theta$:

$$\implies I=\int\frac{2\tan(\theta)\sec^{2}(\theta)}{\tan^{3}(\theta)\sec(\theta)}\mathrm d\theta$$

$$\implies I=2\int\frac{\sec(\theta)}{\tan^{2}(\theta)}\mathrm d\theta$$

Now, write $\sec(\theta)=\frac{1}{\cos(\theta)}$ and $\tan^{2}(\theta)=\frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}$:

$$\implies I=2\int\frac{\cos(\theta)}{\sin^{2}(\theta)}\mathrm d\theta$$

$$\implies I=2\int\csc(\theta)\cot(\theta)\mathrm d\theta$$

$$\implies I=-2\int-\csc(\theta)\cot(\theta)\mathrm d\theta$$

$$\implies I=-2\csc(\theta)+C$$

$$\implies I=-2\sqrt{1+\frac{1}{x}}+C$$

Answer 10

9 years too late!

Since this integral is of the form $\int\frac{\mathrm dx}{L\sqrt Q}$ where $L$ and $Q$ are linear and quadratic functions in $x$ respectively, the substitution $t=\frac1L$ is feasible as on completing the square in the radical in terms of $L$ and taking $L^2$ common from the root, we have $\frac1{L^2}$ in the integrand and a quadratic in $\frac1L$ in the square root.

So, performing the substitution $x=\frac1t$,

$$\begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2+x}}&=\text{sgn}(x)\int\frac{\mathrm dx}{x^2\sqrt{1+\frac1x}}\\&=-\text{sgn}(x)\int\frac{\mathrm dt}{\sqrt{t+1}}\\&=-2\text{sgn}(x)\sqrt{t+1}+C\\&=-2\text{sgn}(x)\sqrt{\frac{x^2+x}{x^2}}+C\\&=\frac{-2\sqrt{x^2+x}}{x}+C\end{align}$$

Answer 11

The given integral is of the form $\displaystyle\int x^m(ax^n+b)^\frac{p}q\mathrm dx$ with $m=-\frac32, n=1, p=-1, q=2, a=1$ and $b=\text{sgn}(x)$. For simplicity, I will consider the case $x>0$. The result for $x<0$ can be derived analogously.

By Chebyshev’s differential binomial, this integral falls into case $(1)$ as $\frac{m+1}n+\frac{p}q=-1\in\mathbb Z$. Hence, performing the substitution $(a+bx^{-n})^\frac1q=\sqrt{1+\dfrac1x}=t$, the integral becomes:

$$\require{cancel}\begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2+x}}&=\int\frac{\mathrm dx}{x^2\sqrt{1+\frac1x}}\\&=\int\cancel{(t^2-1)^2}\cdot\frac1{\cancel t}\cdot\frac{-2\cancel t}{\cancel{(t^2-1)^2}}\mathrm dt\\&=-2t+C\\&=-2\sqrt{1+\frac1x}+C\end{align}$$

Answer 12

Completing the square in the radical gives a hint for a hyperbolic susbtitution, viz., in this case, $x+\frac12=\frac12\cosh t:$

$$\begin{align}\int\frac{\mathrm dx}{x\sqrt{x^2+x}}&=2\int\frac{\mathrm dt}{\cosh t-1}\\&=2\int\frac{\cosh t+1}{\sinh^2t}\mathrm dt\\&=2\int\coth t\text{ csch }t\,\mathrm dt+2\int\text{csch}^2t\,\mathrm dt\\&=-2\text{ csch }t-2\coth t+C\\&=-2\frac{x+1}{\sqrt{x^2+x}}+C\end{align}$$

P.S.: In terms of $t$, the anti-derivative can be concisely written as $-2\coth\dfrac{t}2+C$.