CA Foundation Exam > CA Foundation Questions > ∫ (logx)2 dx and the result isa)x (logx...
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∫ (logx)2 dx and the result is
- a)x (logx)2 – 2 x logx + 2x
- b)x (logx)2 – 2x
- c)2x logx – 2x
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
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∫ (logx)2 dx and the result isa)x (logx)2 – 2 x logx + 2xb...
Integration by Parts:
Integration by parts is a technique used to find the integral of a product of two functions. The formula for integration by parts is given by:
∫u dv = uv - ∫v du
Given Integral:
∫(logx)² dx
Let's choose:
u = logx
dv = logx dx
Calculate the differentials:
du = (1/x) dx
v = ∫logx dx = x logx - x
Apply Integration by Parts:
∫(logx)² dx = logx * (x logx - x) - ∫(x logx - x) * (1/x) dx
∫(logx)² dx = x(logx)² - x² - ∫logx dx
∫(logx)² dx = x(logx)² - x² - x logx + x + C
Simplify the Result:
The final result after simplifying the expression is:
x(logx)² - x² - x logx + x + C
Therefore, the correct answer is option D: none of these.
Integration by parts is a technique used to find the integral of a product of two functions. The formula for integration by parts is given by:
∫u dv = uv - ∫v du
Given Integral:
∫(logx)² dx
Let's choose:
u = logx
dv = logx dx
Calculate the differentials:
du = (1/x) dx
v = ∫logx dx = x logx - x
Apply Integration by Parts:
∫(logx)² dx = logx * (x logx - x) - ∫(x logx - x) * (1/x) dx
∫(logx)² dx = x(logx)² - x² - ∫logx dx
∫(logx)² dx = x(logx)² - x² - x logx + x + C
Simplify the Result:
The final result after simplifying the expression is:
x(logx)² - x² - x logx + x + C
Therefore, the correct answer is option D: none of these.
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∫ (logx)2 dx and the result isa)x (logx)2 – 2 x logx + 2xb)x (logx)2 – 2xc)2x logx – 2xd)none of theseCorrect answer is option 'D'. Can you explain this answer?
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