0
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
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.
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 ).
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.
