∫ln⁡(x+1+x)dx \int \ln ( \sqrt{x+1} + \sqrt{x} ) dx without trigonometry and nice values

Integration:

I watched this video https://www.youtube.com/watch?v=DGpgt8j-nzw in which ∫ln⁡(x+1+x)dx \int \ln ( \sqrt{x+1} + \sqrt{x} ) dx is computed by using trigonometry. Reading the comments, other solutions by using special functions are shown. However, the final antiderivatives do not use any of these functions, so these are intermediate steps in the most literal interpretation of that phrase. I thought of a solution that sidesteps the need of considering any functions other than those present in the integral:

∫ln⁡(x+1+x)dx \int \ln ( \sqrt{x+1} + \sqrt{x} ) dx . Let u=x+1+x u = \sqrt{x+1} + \sqrt{x} . Notice that 1u=x+1−x \frac{1}{u} = \sqrt{x+1} - \sqrt{x} . Then du=12(1x+1+1x)dx=12ux+1xdx du = \frac{1}{2} \left( \frac{1}{\sqrt{x+1}} + \frac{1}{\sqrt{x}} \right)dx = \frac{1}{2} \frac{u}{\sqrt{x+1} \sqrt{x}}dx . To write everything in terms of u u notice that u2−1u2=4x+1x  ⟹  2x+1x=u4−12u2 u^2 - \frac{1}{u^2} = 4 \sqrt{x+1} \sqrt{x} \implies 2 \sqrt{x+1} \sqrt{x} = \frac{u^4 - 1}{2u^2} . Thus du=uu4−12u2dx=2u3u4−1dx du = \frac{u}{\frac{u^4 - 1}{2u^2}}dx = \frac{2u^3}{u^4 - 1} dx . Going back to the original integral:

∫ln⁡(x+1+x)dx=∫ln⁡u2u3(u4−1)dx=∫u2ln⁡u−ln⁡u2u3du \int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \int \frac{\ln u}{2u^3} (u^4 - 1)dx = \int \frac{u}{2} \ln u - \frac{ \ln u}{2u^3} du . Integrating the terms in the integral is now a standard exercise in integration by parts which I will skip. The result is 18u2(2ln⁡u−1)+2log⁡u+18u2 \frac{1}{8}u^2(2 \ln u - 1) + \frac{2 \log u + 1}{8u^2} . Giving a final result of:

∫ln⁡(x+1+x)dx=18(x+1+x)2(2ln⁡(x+1+x)−1)+2ln⁡(x+1+x)+18(x+1+x)2+C \int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{8}(\sqrt{x+1} + \sqrt{x})^2(2 \ln (\sqrt{x+1} + \sqrt{x}) - 1) + \frac{2 \ln (\sqrt{x+1} + \sqrt{x}) + 1}{8(\sqrt{x+1} + \sqrt{x})^2} + C

Nice values:

From the form of the antiderivative we can expect this integral to give us nice values whenever x+1+x \sqrt{x+1} + \sqrt{x} is a nice power of e e . The equation x+1+x=ek \sqrt{x+1} + \sqrt{x} = e^k has solution x=14e−2k(e2k−1)2 x= \frac{1}{4} e^{-2 k} (e^{2 k} - 1)^2 . You can obtain this by converting the equation into a normal quadratic equation. The nicest value I was able to find was taking k=0 k=0 to yield x=0 x = 0 and k=12 k = \frac{1}{2} to yield x=(e−1)24e x = \frac{(e-1)^2}{4e} . Then: ∫0(e−1)24eln⁡(x+1+x)dx=14e \int_{0}^{\frac{(e-1)^2}{4e}} \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{4e}

#Calculus

Note by Leonel Castillo
2 years, 10 months ago

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Comments

Here's another way to solve it:

∫ln⁡(x+1+x) dx=12∫ln⁡(x+1+x)2 dx=12∫ln⁡(2x+1+2x2+x) dx=12∫ln⁡[2(x+12)+2(x+12)2−14  ] dx \begin{aligned} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln\left(\sqrt {x+1} + \sqrt x\right)^2 \, dx \\ &=& \frac12 \int \ln\left(2x + 1 + 2\sqrt{x^2 + x}\right) \, dx \\ &=& \frac12 \int \ln\left[ 2\left(x+ \frac12\right) + 2\sqrt{\left(x + \frac12\right)^2 - \frac14} \; \right] \, dx \\ \end{aligned}

Let x+12=12sec⁡θx + \frac12 = \frac12 \sec \theta, then (x+12)2−14=(12tan⁡θ)2 \left(x+\frac12\right)^2 - \frac14 = \left(\frac12 \tan\theta\right)^2 and dxdθ=12sec⁡θtan⁡θ \frac{dx}{d\theta} = \frac12\sec\theta\tan\theta. The integral becomes

∫ln⁡(x+1+x) dx=12∫ln⁡(sec⁡θ+tan⁡θ)⋅12sec⁡θtan⁡θ dθ=14∫ln⁡(sec⁡θ+tan⁡θ)⏟=u⋅sec⁡θtan⁡θ dθ⏟=dv, Integrate by parts=14(uv−∫v du)=14[ln⁡(sec⁡θ+tan⁡θ)sec⁡θ−∫sec⁡θ⋅sec⁡θtan⁡θ+sec⁡2θsec⁡θ+tan⁡θ dθ]=14[sec⁡θln⁡(sec⁡θ+tan⁡θ)−∫sec⁡2θ dθ]=14[sec⁡θln⁡(sec⁡θ+tan⁡θ)−tan⁡θ]+C=14[(2x+1)ln⁡((2x+1)+(2x+1)2−1  )−(2x+1)2−1  ]+C=2x+14ln⁡(2x+1+2x2+x  )−12x2+x+C=2x+12ln⁡(x+1+x)−12x2+x+C \begin{aligned} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln (\sec \theta + \tan\theta) \cdot \frac12 \sec \theta\tan\theta \, d\theta \\ &=& \frac14 \int \underbrace{\ln (\sec \theta + \tan\theta)}_{=u} \cdot \underbrace{\sec \theta\tan\theta \, d\theta}_{=dv} , \qquad\qquad \text{ Integrate by parts} \\ &=& \frac14 \left (uv - \int v \, du\right) \\ &=& \frac14 \left [ \ln (\sec \theta + \tan\theta) \sec \theta - \int \sec \theta \cdot \dfrac{\sec\theta \tan\theta + \sec^2\theta}{\sec \theta + \tan \theta} \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \int \sec^2\theta \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \tan \theta \right ] +C \\ &=& \frac14 \left [(2x + 1) \ln \left ((2x+1) + \sqrt{(2x+1)^2-1}\; \right) - \sqrt{(2x+1)^2-1} \; \right ] +C \\ &=& \frac{2x+1}4 \ln \left (2x+1 + 2 \sqrt{x^2+x}\; \right) - \frac12 \sqrt{x^2+x} +C \\ &=& \frac{2x+1}2 \ln \left (\sqrt {x+1} + \sqrt x\right) - \frac12 \sqrt{x^2+x} +C \\ \end{aligned}

which is identical to blackpenredpen's final answer.

Pi Han Goh - 2 years, 10 months ago

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Pi Han Goh, your post does not apply, because the OP expressly asked for the solution to be done without using trigonometry.

Linda Slovik - 2 years, 10 months ago

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I did not ask, I was just sharing a solution. If he wants to use this post to post another alternative solution that's okay.

Leonel Castillo - 2 years, 10 months ago

OP, your answer is wrong, because you do not include the "+ C."

Linda Slovik - 2 years, 10 months ago

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Thank you.

Leonel Castillo - 2 years, 10 months ago

Just a thought- How many different sums can be made by adding at least two different numbers from the set of integers {5,6,7,8}?

Lucia and Emma - 2 years, 8 months ago
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