Integral Calculus Question
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Integral Calculus Question

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
Just get the antiderivative,So you get 4(1/3)(x^3) - 5x + C-int(x/(4*x^2-5), x) = (1/8)*ln(4*x^2 - 5) + C-This is standard high school calculus. u = 4x^2 - 5 so du/dx = 8x. Substitute it and then use the integration operator. Anyone can do it.......
What is the answer to this given equation using 'u' substitution.

∫ (x) / (4x^2 - 5) dx

I found this question on the 'net but suspected the final answer to be wrong.

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∫(x)/(4x² - 5) dx

u = 4x² - 5

du/8 = x dx

1/8*∫du/u

1/8*ln|4x² - 5| + C

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You don't need "u" substitution for this equation.

Just get the antiderivative, is 4(x^(2 + 1))/(2 + 1) - 5x + constant

So you get 4(1/3)(x^3) - 5x + C

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int(x/(4*x^2-5), x) = (1/8)*ln(4*x^2 - 5) + C

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This is standard high school calculus. u = 4x^2 - 5 so du/dx = 8x. Substitute it and then use the integration operator. Anyone can do it.
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