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How do you integrate #int (2x-sinx) / (x^2 + cosx)#?

Answer 1

Aurora Alvarado

The numerator is the derivative of the denominator.

Use substitution:

Let #u=x^2+cosx#, which makes #du = (2x-sinx) dx#

So the integral is #int (du)/u = ln absu +C#.

#int (2x-sinx)/(x^2+cosx) dx = ln abs(x^2+cosx) +C#

# = ln(x^2+cosx)+C# #" "# (since #x^2+cosx > 0# for all #x#)

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Answer 2

Elijah Abram

To integrate (\frac{{2x - \sin(x)}}{{x^2 + \cos(x)}}), you can use substitution method. Let (u = x^2 + \cos(x)), then (du = (2x - \sin(x))dx). The integral becomes (\int \frac{1}{u} du), which is (\ln|u| + C). Substitute back (u = x^2 + \cos(x)) to get the final answer: (\ln|x^2 + \cos(x)| + C).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.