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(x^2+4x+10)/(x+3)(x+4)(x+1) = A/(x+3) + B/(x+4) + C/(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)]/(x+3)(x+4)(x+1) = (x^2+4x+10)/(x+3)(x+4)(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)] = (x^2+4x+10) A(...
Differentiate the equationdy/dx=6x+2x(dy/dx)+2y-4y(dy/dx)Set this equal to zero and solve for dy/dx which gives:dy/dx=(2y+6x)/(4y-2x)For x=2 and y=4 dy/dx=5/3(y-yo)=m(x-xo)y-4=(5/3)(...
3x3 /3 + 4x6/(6x3) = x3 + 2x6/9 + c
Firstly, ensure both numerator and denominator are factorised fully so that any possible cancellations can happen - making it easy for yourself is always a good idea. Now, turn it into partial fractions. ...
Firstly we notice that the numerator is the derivative of the denominator so we can use integration by subsitution method. Setting u=x^(2)+1. We can differentiate this to get du/dx=2x Subbing in dx=du/2x ...
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