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Integrate square root of 4 xdivide by x?
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
Integration of ln 5x dx?
5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c
How do you integrate x power -1?
x-1 = 1/x ∫1/x dx = ln x + C
How do you integrate piecewise continuous functions?
Simply integrate all the pieces apart, en add them up. This is allowed, because int_a^c f(x)dx = int_a^b f(x)dx + int_b^c f(x)dx for all a,b,c in dom(f).
Integration of x ln 5x dx?
int x ln5x dx by parts u = ln5x du = 1/5x or 5x^-1 dv = x v = 1/2x^2 uv - int v du ln5x 1/2x^2 - int 1/2x^2 5x^-1 1/2ln5x*x^2 - 1/6x^3 5x + C
What is the derivative of 2 to the power of 5x?
25x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(25x)=25x*ln(2)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(25x)=95x*ln(2)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(25x)=25x*ln(2)*(5*1)d/dx(25x)=25x*ln(2)*(5)-25x can simplify to (25)x, which equals 32x.d/dx(95x)=32x*ln(2)*(5)
How do you Integrate the following S4x3-2x2 x-1dx in Calculus?
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
How do you find the derivative of 9 to the 5x?
95x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(95x)=95x*ln(9)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(95x)=95x*ln(9)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(95x)=95x*ln(9)*(5*1)d/dx(95x)=95x*ln(9)*(5)-95x can simplify to (95)x, which equals 59049x.-ln(9) can simplify to ln(32), so you can take out the exponent to have 2ln(3).d/dx(95x)=59049x*2ln(3)*(5)d/dx(95x)=10*59049x*ln(3)
How do you integrate cotxdx?
The integral of cot(x)dx is ln|sin(x)| + C
How you integrate xxxx 1 dx?
.2x^5+x+C
Integrate square root of 4 xdivide by x?
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
Integration of ln 5x dx?
5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c
How do you integrate x power -1?
x-1 = 1/x ∫1/x dx = ln x + C
How do you integrate piecewise continuous functions?
Simply integrate all the pieces apart, en add them up. This is allowed, because int_a^c f(x)dx = int_a^b f(x)dx + int_b^c f(x)dx for all a,b,c in dom(f).
Integrate of ln x squared?
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
Integration of x ln 5x dx?
int x ln5x dx by parts u = ln5x du = 1/5x or 5x^-1 dv = x v = 1/2x^2 uv - int v du ln5x 1/2x^2 - int 1/2x^2 5x^-1 1/2ln5x*x^2 - 1/6x^3 5x + C
How do you integrate square root of 1 over x-1 dx?
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