Integrals with $2$ different radicals in them such as $\int \frac{\sqrt{x}}{x^2\left(\sqrt{x+1}+\sqrt{x}\right)}\mathrm dx$

Question

I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:

$1.$ $$\int \frac{\sqrt{x}}{x^2\left(\sqrt{x+1}+\sqrt{x}\right)}\mathrm dx$$

$2.$ $$\int \frac{\sqrt[3]{x+2}-\sqrt[3]{x}}{x^2\left(\sqrt[3]{x+2}+\sqrt[3]{x}\right)}\mathrm dx$$

My attempt for the first integral:

\begin{align} \int \frac{\sqrt{x}}{x^2\left(\sqrt{x+1}+\sqrt{x}\right)}\mathrm dx &= \int \frac{\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right)}{x^2\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\mathrm dx \\ &= \int \frac{\sqrt{x}\sqrt{x+1}-x}{x^2}\mathrm dx \end{align}

Now, I can split it into two integrals. The problem is with:

$$\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}\mathrm dx$$

and the major problem is that I don't know how to evaluate integrals that have some distinct roots with different values inside those roots. The second integral seems even harder.

If speaking of "different values under roots", I am only familiar with evaluating such integrals:

$$\int \frac{\left(\sqrt{\frac{x+2}{x-1}}-1\right)^2}{3\left(\sqrt{\frac{x+2}{x-1}}+2\right)}\mathrm dx$$ because there's a simple algorithm that I can follow to evaluate it.

Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.

EDIT: \begin{align}\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}\mathrm dx&= \int \frac{\sqrt{x^2+x}}{x^2}\mathrm dx\\&= \int \frac{x^2 + x }{x^2\sqrt{x^2+x}}\mathrm dx\\&= \int \frac{x+1}{x\sqrt{x^2+x}}\mathrm dx\end{align}

And now, Euler's substitution should work.

Answer 1

Hint: For the integral $$\int \frac{\sqrt{x^2+x}}{x^2}dx$$ substitute $$\sqrt{x^2+x}=x+t$$ it is the Eulerian substitution. Then we get by squaring $$x=\frac{t^2}{1-2t}$$ and $$dx=-2\,{\frac {t \left( -1+t \right) }{ \left( -1+2\,t \right) ^{2}}}dt$$

Answer 2

$1.$ The integral you got stuck at can be evaluated very easily by rationalizing the numerator:

$$\begin{align}\int\frac{\sqrt{x^2+x}}{x^2}\mathrm dx&=\int\frac{\mathrm dx}{\sqrt{x^2+x}}+\int\frac{\mathrm dx}{x\sqrt{x^2+x}}\\&=\int\frac{\mathrm dx}{\sqrt{\left(x+\frac12\right)^2-\frac14}}+\text{sgn}(x)\int\frac{\mathrm dx}{x^2\sqrt{1+\frac1x}}\end{align}$$

You can take it from here.

$2.$ The integral can be simplified by substituting $t=\frac1x$, then rationalizing the denominator:

$$\begin{align}\int\frac{\sqrt[3]{x+2}-\sqrt[3]x}{x^2(\sqrt[3]{x+2}+\sqrt[3]x)}\mathrm dx&=\int\frac{1-\sqrt[3]{2t+1}}{1+\sqrt[3]{2t+1}}\mathrm dt\\&=\int\frac{(1-\sqrt[3]{2t+1})(1+(2t+1)^\frac23-\sqrt[3]{2t+1})}{2t+2}\mathrm dt\\&\overset{y=t+1}{=}-\int\frac{t}{t+1}\mathrm dt-\int\frac{\sqrt[3]{2y-1}}y\mathrm dy+\int\frac{(2y-1)^\frac23}y\mathrm dy\end{align}$$

Note that the second and the third integrals are of the form of Chebyshev’s differential binomial, namely $\frac{a+1}b\in\mathbb Z$ implying that they can be evaluated by substituting $2y-1=u^3$. After this, you’ll end up with the integrals of rational functions(with denominator $u^3-1$) that can be evaluated by PFD.

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A simpler way to evaluate this integral after substituting $t=\frac1x$ is to split it as follows followed by the substitution $2t+1=u^3$:

$$\begin{align}\int\frac{\sqrt[3]{x+2}-\sqrt[3]x}{x^2(\sqrt[3]{x+2}+\sqrt[3]x)}\mathrm dx&=\int\frac{1-\sqrt[3]{2t+1}}{1+\sqrt[3]{2t+1}}\mathrm dt\\&=t-2\int\frac{\mathrm dt}{\sqrt[3]{2t+1}+1}\\&=t-3\int\frac{u^2}{u+1}\mathrm du\end{align}$$

which can now be finished off by partial fractions.

P.S. For the first integral, you shouldn’t write $\sqrt{x^2+x}=\sqrt x\sqrt{x+1}$ if you want to evaluate it in general, i.e., this is true only for $x>0$, whereas the domain of the integrand is $(-\infty,-1]\cup(0,\infty)$.

Answer 3

ALITER

$$I=\int \frac{\sqrt{x}}{x^2 \left(\sqrt{1+x}+\sqrt{x}\right)}dx=\int \frac{dx}{x^2 \left(\sqrt{1+\frac{1}{x}}+1\right)}$$

$$1+\frac{1}{x}\longrightarrow t$$

$$I=-\int \frac{dt}{\sqrt{t}+1}= \ -\int\frac{\sqrt{t}-1}{t-1}dt = \int\frac{dt}{t-1}-\int \frac{\sqrt{t}}{t-1}dt$$

$$I_{1}=\int\frac{dt}{t-1}=\ln|t-1|+C$$

$$I_{2}=\int\frac{\sqrt{t}}{t-1}dt=\int\frac{t}{\sqrt{t}(t-1)}dt$$ $$\sqrt{t}\longrightarrow u$$

$$I_{2}=\int\frac{2u^2}{u^2-1}du=\int2du \ \ + \ \int\frac{2}{u^2-1}du \ = \ 2u \ + \ \ln|1-u| \ - \ \ln|1+u| \ + \ C$$

$$I_{2} \ = \ 2\sqrt{t} \ + \ \ln|1-\sqrt{t}|\ - \ln|1+\sqrt{t}|+C $$

And finally, $$\boxed{I\ =\ I_{1}-I_{2} \ =\ 2\ln|1+\sqrt{t}| \ - 2\sqrt{t} \ + \ C} $$

Which agrees with Wolfram Alpha's Answer

Answer 4

I will solve this integral by Trigonometric Substitution process.

$1)$ Let $$I=\int\frac{\sqrt{x}}{x^{2}(\sqrt{x+1}+\sqrt{x})}dx$$ Then

$$I=\int\frac{\sqrt{x}(\sqrt{x+1}-\sqrt{x})}{x^{2}[(x+1)-(x)]}dx=\int\frac{\sqrt{x}\sqrt{x+1}-x}{x^{2}}dx \int\frac{\sqrt{x}\sqrt{x+1}}{x^{2}}dx-\int\frac{1}{x}dx$$

Now let's assume that $$I_{1}=\int\frac{\sqrt{x}\sqrt{x+1}}{x^{2}}dx$$

And $$I_{2}=\int\frac{1}{x}dx$$

Now

$I_{1}=\int\frac{\sqrt{x}\sqrt{x+1}}{x^{2}}dx$

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

\begin{align} I_{1} &=2\int\frac{\tan(\theta)\sec(\theta)\tan(\theta)\sec^{2}(\theta)}{\tan^{4}(\theta)}d\theta=2\int\frac{\sec^{3}(\theta)}{\tan^{2}(\theta)}d\theta\\ &=2\int\frac{1}{\cos(\theta)\sin^{2}(\theta)}d\theta =2\int\frac{\cos(\theta)}{\cos^{2}(\theta)\sin^{2}(\theta)}d\theta \end{align} Now let's assume that $\sin(\theta)=z$ which implies $\cos(\theta)d\theta=dz$

$\implies I_{1}=2\int\frac{1}{(1-z^{2})z^{2}}dz$

Hope $I_{1}$ will be done now.

And $$I_{2}=\int\frac{1}{x}dx$$

$\implies I_{1}=\ln|x|+C$

Answer 5

For (1), Euler's Third Substitution, $\sqrt{x^2 + x} = x t$, rationalizes the integral, giving $$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}\,dx = 2 \int \left(\frac1{1 + t} - 1\right) \,dt .$$

For (2), rewriting the given integral as $$\int \frac{\sqrt[3]{1 + \frac2x} - 1}{\sqrt[3]{1 + \frac2x} + 1} \cdot \frac{dx}{x^2}$$ suggests the rationalizing substitution $u^3 = 1 + \frac2x$, $3 u^2 \,du = -\frac{2 \,dx}{x^2}$, which yields $$\frac32 \int \frac{u^2 (1 - u)}{1 + u} \,du = \frac32 \int \left(u^2 - 2 u + 2 - \frac2{1 + u}\right) du .$$