I = ∫sqrt(4x^2+20) dx I = ∫(1)/(x^2*sqrt(x^2-a)) dx

I = ∫sqrt(4x^2+20) dx

I = ∫(1)/(x^2*sqrt(x^2-a)) dx

Expert Solution

milcah answered 2 years ago

integrate from infinity to 1: (1+(sin^2(x))/sqrt (x)) dx

((sqrtx)+1)dy/dx=(ysqrtx)/(x) + sqrt(y/x), y(2)=1

Integral x^5-x^4+4x^3-4x^2+8x-4 / ( x^3+2)^3(x^2-1) dx

Evaluate each integral using trig substitutions 1.) Integral of (3x^5dx)/(sqrt(16-x^2) 2.) Integral of (sqrt(x^2-16)dx)/x 3.) Integral...

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1.) (10pts) Consider the following differential equation: (x^2)(dy/dx)=2x(sqrt(y))+(x^3)(sqrt(y)) a)Determine whether the equation is separable (S), linear...

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I Integrate x^2 e^(-.5(x+1))^2 dx from -infinity to +infinity

Find dy/dx by implicit differentiation. 11. ycosx=x^2+y^2 13. sqrt(x+y)=x^4+y^4 15. tan(x/y)=x+y 17. sqrt(xy)=1+(x^2)*(y) 19. sin(xy) =...

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solve for x: [x * sqrt(1+x2)] + ln[x + sqrt(1+x2)] = 25

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