- #1
DaalChawal
- 85
- 0
Integration Doubt: Answers & Solutions
- Context: MHB
- Thread starter DaalChawal
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- Doubt Integration
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SUMMARY
The discussion focuses on simplifying the integrand of a complex logarithmic expression involving variables and their logarithms. The expression is transformed to a more manageable form, highlighting the use of properties of logarithms and algebraic manipulation. Participants explore differentiation techniques and substitution methods to further analyze the integral, although they encounter challenges in progressing beyond the simplification stage.
PREREQUISITES- Understanding of logarithmic properties and identities
- Familiarity with integration techniques in calculus
- Knowledge of differentiation and substitution methods
- Basic algebraic manipulation skills
- Study advanced integration techniques, focusing on logarithmic integrals
- Learn about substitution methods in calculus for simplifying integrals
- Explore differentiation of logarithmic functions and their applications
- Investigate properties of exponential functions in relation to logarithms
Students and professionals in mathematics, particularly those studying calculus, integrals, and logarithmic functions. This discussion is beneficial for anyone looking to enhance their problem-solving skills in advanced mathematical contexts.
- #2
Prove It
Gold Member
MHB
- 1,434
- 20
It might be worth simplifying the integrand...
$\displaystyle \begin{align*} \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }} &= \frac{ \ln{\left( \mathrm{e} \right) } + \ln{ \left( x^{x+1} \right) } + \left[ \sqrt{x} \, \ln{ \left( x \right) } \right] ^2 }{ 1 + x\ln{ \left( x \right) } \left[ \ln{\left( \mathrm{e}^x \right) } + \ln{ \left( x^x \right) } \right] } \\ &= \frac{ 1 + \left( x + 1 \right) \ln{ \left( x \right) } + x \, \left[ \ln{ \left( x \right) } \right] ^2 }{ 1 + x \ln{ \left( x \right) } \left[ x + x\ln{ \left( x \right) } \right] } \end{align*}$
I don't know if this helps, but it looks simpler at least...
$\displaystyle \begin{align*} \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }} &= \frac{ \ln{\left( \mathrm{e} \right) } + \ln{ \left( x^{x+1} \right) } + \left[ \sqrt{x} \, \ln{ \left( x \right) } \right] ^2 }{ 1 + x\ln{ \left( x \right) } \left[ \ln{\left( \mathrm{e}^x \right) } + \ln{ \left( x^x \right) } \right] } \\ &= \frac{ 1 + \left( x + 1 \right) \ln{ \left( x \right) } + x \, \left[ \ln{ \left( x \right) } \right] ^2 }{ 1 + x \ln{ \left( x \right) } \left[ x + x\ln{ \left( x \right) } \right] } \end{align*}$
I don't know if this helps, but it looks simpler at least...
- #3
DaalChawal
- 85
- 0
I also reached this step I tried to create differentiation inside and used substitution too but could not solve it further.
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