We have $t>0$. Recall that $$\int\frac1{(t^2-x^2)^{1/2}}\ dx=\arcsin\left(\frac xt\right)+c_1$$ Take the derivative of both sides with respect to $t$ to get $$\int\frac{-t}{(t^2-x^2)^{3/2}}\ dx=\frac{-x}{|t|\sqrt{t^2-x^2}}+c_2$$ Setting $t=1$ and simplifying, we reach $$\int\frac1{(1-x^2)^{...

You could try differentiation under the integral sign: $$\frac d{dt}\int F(xt)\ dx\bigg|_{t=1}=\int xf(x)\ dx$$ So you'll need to solve $\int F(x)\ dx$ to continue further.

We have $t>0$. Recall that $$\int\frac1{(t^2-x^2)^{1/2}}\ dx=\arcsin\left(\frac xt\right)+c_1$$ Take the derivative of both sides with respect to $t$ to get $$\int\frac{-t}{(t^2-x^2)^{3/2}}\ dx=\frac{-x}{|t|\sqrt{t^2-x^2}}+c_2$$ Setting $t=1$ and simplifying, we reach $$\int\frac1{(1-x^2)^{...

I was recently looking through integration techniques when I came upon differentiation under the integral sign (DUIS). It seems to be pretty powerful, for example: $$f(t)=\int_0^1\frac{x^t-1}{\ln(x)}\ dx\implies f'(t)=\int_0^1x^t\ dx=\frac1{t+1}\\\implies f(t)=C+\ln(t+1)\\f(0)=0\implies C=0\\\i...

The integral itself is: $$\int \frac{x^4-2x^2+2}{x^3-2x^2-x+2}dx$$ After long division I got: $$\int \Big(x+2+\frac{3x^2-2}{x^3-2x^2-x+2}\Big)dx$$ And after simplifying the denumerator I got: $$x^3-2x^2-x+2 = x^2(x-2)-1(x-2) = (x^2-1)(x-2)$$ But I am not sure about $A$ and $B$ values Shoul...

Interesting problem! Let's start with the following integral: $$f(t)=\int_0^1e^{(2x-1)t}e^{x-x^2}\ dx=\frac12\sqrt\pi e^{t^2+\frac14}\left[\operatorname{erf}\left(\frac12-t\right)+\operatorname{erf}\left(\frac12+t\right)\right]$$ It follows obviously that $$I_0=f(0)$$ $$I_1=f'(0)$$ etc. ...

Note that it says on Wikipedia that [...] in Euclidean geometry, a point is a primitive notion upon which the geometry is built. Being a primitive notion means that a point cannot be defined in terms of previously defined objects. That is, a point is defined only by some properties, called a...