2016年12月21日 · The integrand is defined only for #x in (-1,1)# so we can substitute #x=sint#, #dx=costdt# with #t in [-pi/2,pi/2]#

2017年1月30日 · How do you evaluate the integral #int 1/(xsqrt(1-x^2))#? Calculus Techniques of Integration Integration by Parts. 1 Answer

2017年1月11日 · #int (dx)/(xsqrt(1+x^2)) = 1/2ln (sqrt(1+x^2) -1) -1/2 ln (sqrt(1+x^2)+1)+C# where we have dropped the absolute value, because the arguments of the logarithms are always positive. Applying the properties of logarithms this can also be written as:

2015年8月17日 · The solutions are color(blue)(x=-1,x=1/2 2x^2+x−1=0 We can Split the Middle Term of this expression to factorise it.

2016年6月1日 · How do you write the partial fraction decomposition of the rational expression # (3x^2 + 12x

2014年9月25日 · You find the zeros by setting the expression equal to zero. x^2+1=0 Then solve for x x^2=-1 sqrt(x^2)=+-sqrt(-1) x=+-i This expression has 2 imaginary roots +i and -i.

2018年1月2日 · So, our Maclaurin series is sum_(n=0)^oox^(2n)=1+x^2+x^4+x^6... We know that the geometric series is convergent for |x|<1, so to determine when this series is convergent, we solve for when |x^2|<1: |x^2|<1 x^2 is always positive, so we don't need the absolute value: x^2<1 sqrt(x^2)< sqrt1 |x|<1 As luck would have it, this series has the same ...

2015年7月1日 · The remainder when dividing f(x) by (x+2)(x-1) is (7x-5) Note that (x+2)(x-1) = x^2+x-2 is quadratic, so if we divide another polynomial by it then the remainder will be 0, a non-zero constant or a linear polynomial. Any remainder of greater degree could be divided further.

2017年5月10日 · How to you find the general solution of #dy/dx=x/(1+x^2)#? Calculus Applications of Definite Integrals Solving Separable Differential Equations

2018年5月21日 · 15942 views around the world You can reuse this answer ...