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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
What else can I help you with?
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
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the derivative of ln 1 divided by 1-x?
0
Integration of tangent x?
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
What is the derivative of sin x to the e to the xth power?
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
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 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)
Integral of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
What is the anti-derivative of ln x?
The anti-derivative of ( \ln x ) can be found using integration by parts. Let ( u = \ln x ) and ( dv = dx ), then ( du = \frac{1}{x} dx ) and ( v = x ). Applying integration by parts, we get: [ \int \ln x , dx = x \ln x - \int x \cdot \frac{1}{x} , dx = x \ln x - x + C, ] where ( C ) is the constant of integration. Thus, the anti-derivative of ( \ln x ) is ( x \ln x - x + C ).
Integration by parts of x tanx?
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
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 antiderivaative of xlnx?
To find the antiderivative of ( x \ln x ), we can use integration by parts. Let ( u = \ln x ) and ( dv = x , dx ), which gives ( du = \frac{1}{x} , dx ) and ( v = \frac{x^2}{2} ). Applying integration by parts, we get: [ \int x \ln x , dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x , dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C, ] where ( C ) is the constant of integration. Thus, the antiderivative of ( x \ln x ) is ( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C ).
Integration of xtanx?
To integrate ( x \tan(x) ), we can use integration by parts. Let ( u = x ) and ( dv = \tan(x) , dx ). This gives ( du = dx ) and ( v = -\ln|\cos(x)| ). Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we obtain: [ \int x \tan(x) , dx = -x \ln|\cos(x)| + \int \ln|\cos(x)| , dx + C ] The integral ( \int \ln|\cos(x)| , dx ) does not have a simple closed form, so the final result may be expressed in terms of this integral along with the logarithmic term.
How do you integrate x secx?
To integrate ( x \sec x ), you can use integration by parts. Let ( u = x ) and ( dv = \sec x , dx ). Then, ( du = dx ) and ( v = \ln |\sec x + \tan x| ). Applying the integration by parts formula, you get: [ \int x \sec x , dx = x \ln |\sec x + \tan x| - \int \ln |\sec x + \tan x| , dx + C ] where ( C ) is the constant of integration. The second integral may require further techniques to evaluate.
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the integral of a constant to the power of x with respect to x?
∫ ax dx = ax/ln(a) + C C is the constant of integration.
What is the integral of the cotangent of x with respect to x?
∫ cot(x) dx = ln(sin(x)) + CC is the constant of integration.
