6
votes
2answers
145 views

A strange answer for $\int _{-1}^1 \log x\; dx$

I typed $\int _{-1}^1 \log x\; dx$ on Wolfram Alpha. It is giving the answer to be $-2+i\pi$. Can someone please explain what is happening?
1
vote
1answer
146 views

Evaluation of the integral $\int_{-6}^{-3}\frac{\sqrt{x^2-9}}{x}$

How to evaluate the following integral? I have tried the following things but I have no idea to continue after the last step. Moverover, the integral seems wrong when compared with the ans from ...
1
vote
1answer
42 views

Please explain the steps in the solution of this integral

After unsuccessfully solving an integral, I turned to WRA for some guidance. Unfortunately, I don't understand what happen in the solution. More particularly, I don't understand why the numerator has ...
0
votes
1answer
95 views

Why WolframAlpha gives different results?

Any idea why WolframAlpha gives different answers when computing the indefinite integral of these functions with respect to x first or with respect to y first? ...
1
vote
2answers
101 views

How can these two be equivalent (wolfram-alpha incorrect) !?

So wolfram-alpha reads The integral of $$\int \frac{1}{\sqrt{a^2-x^2}}dx=\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$$ but that $$\int\frac{1}{\sqrt{a^2-x^2}}dx \;\mathrm{where}\; a=5 ...
2
votes
1answer
59 views

Wolfram-Alpha's choice of $k$ in a complex logarithm

I'm puzzling on this complex integral: $$ \int \frac{2ie^{it}}{2e^{it} - 1}dt = \log(2e^{it} -1)$$ The numerator is the derivative of the divisor, so the primitive is the log of the divisor. When ...
3
votes
2answers
120 views

Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$

I calculated this integral : $$\int{x \cdot \sin^2(x) dx}$$ By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a ...
1
vote
2answers
575 views

Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$

I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?
1
vote
1answer
266 views

Moving the negative out of the integral - a bug in Wolfram Alpha?

A guy I follow on Twitter seems to be having some issues with Wolfram Alpha - namely that moving a negative outside of an integral results in two different answers: $\int -f(x) = -\int f(x)$ gives ...
4
votes
1answer
5k views

Contour integration and wolfram

I am not sure if this is the right site to ask this question, but here it is: Does Wolfram Alpha do contour integrals? If so how might I access that feature. Also, is there an online manual for ...
1
vote
1answer
112 views

An integral evaluation

I tried my luck with Wolfram Alpha, with $p \in \mathbb{R}$ $$\int_{-\infty}^{\infty} \frac{x^p}{1+x^2} dx = \frac{1}{2} \pi ((-1)^p+1) \sec(\frac{\pi p}{2})$$ for $-1<p<1$, and doesn't exist ...
0
votes
1answer
114 views

Help me with this mock-long-division $\frac{-x+2}{x^{2}+x-2}=\frac{-4}{3(x+2)}+\frac{1}{3(x-1)}$

$$\frac{-x+2}{x^{2}+x-2}=\frac{-4}{3(x+2)}+\frac{1}{3(x-1)}$$ Wolphram Alpha states that one can do this with long division, I cannot immediately realize it. Could someone show the trick to simplify ...
2
votes
2answers
1k views

Wolfram Alpha: How to show functions with abs() and sgn() as a piecewise function?

I was trying to obtain this result: integrate e^|x| dx The result is $$ \int e^{|x|} dx = \frac12 e^{-x} \left( (e^x-1)^2 \operatorname{sgn}(x) -2e^x +e^{2x} - ...