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sumerman
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i have tried some methods like uv - integral vdu but can't reach the answer
tony.c.tan said:Try to take the derivative of this with respect to $x$, and see what do you get:
[tex]e^x\left[\ln x-\sum_{i=1}^{\infty}(i-1)!x^{-i}\right][/tex]
tony.c.tan said:Try to take the derivative of this with respect to $x$, and see what do you get:
[tex]e^x\left[\ln x-\sum_{i=1}^{\infty}(i-1)!x^{-i}\right][/tex]
g_edgar said:Woh! a series that diverges for every x ... what a useful answer ...
norice4u said:apologize for my ignorance but what is that process called and the Sigma looking symbol?
NB: excuse me for fail to type with mathematic symbol
Anti-Differentiate x^x=?
But really i am asking how to anti-differentiate x^x(lnx+1) which comes from the derivative of y=x^x
Because out of curiosity i always hold the belief in math if there is a forward operation there should be a backwards operation so if i can differentiate x^x to get that ugly function to anti-differentitate what operations would i have to undergo.
The general formula for calculating the integral of e^x.(lnx)dx is ∫e^x.lnx dx = e^x(lnx-1)+C, where C is the constant of integration.
To solve the integral of e^x.(lnx)dx using integration by parts, we first identify u=e^x and dv=lnx dx. Then, we use the formula ∫u dv = uv - ∫v du to get the final answer of e^x(lnx-1)-∫e^x/x dx.
Yes, we can use the substitution method by letting u=lnx. This will result in the integral becoming ∫e^(e^u) du, which can be solved using the power rule for integration.
Yes, the u-substitution method can be used to solve the integral of e^x.(lnx)dx. By letting u=lnx, the integral becomes ∫e^(e^u) du, which can then be solved using the power rule for integration.
Yes, we can use the definite integrals property to solve the integral of e^x.(lnx)dx. By using the property ∫a^b f(x) dx = F(b)-F(a), where F(x) is the antiderivative of f(x), we can evaluate the definite integral by plugging in the limits of integration into the antiderivative formula.