(1/4)x^2(2logx-1) + c
d/dx ( log x) = 1/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 ).
-logx=21.1 logx=-21.1 e^-21.1=x
∫ tan(x) dx = -ln(cos(x)) + C C is the constant of integration.
∫ cosh(x) dx = sinh(x) + C C is the constant of integration.
d/dx ( log x) = 1/x
X(logX-1) + C
The integral of f'(x) = 1 is f(x) = x + c
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 ).
logx^2=2 2logx=2 logx=1 10^1=x x=10
∫ d/dx f(x) dx = f(x) + C C is the constant of integration.
-logx=21.1 logx=-21.1 e^-21.1=x
y=logx y=10 logx= 10 10logx = 10log1 logx = log1 x = 1 //NajN
∫ sin(x) dx = -cos(x) + CC is the constant of integration.
∫ cos(x) dx = sin(x) + CC is the constant of integration.
∫ cot(x) dx = ln(sin(x)) + CC is the constant of integration.
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 -----------------------------------------