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Find the value $C$ such that $\int^{\infty}_{0}(\frac{x}{x^2+1} - \frac{C}{3x+1})dx$ converges

$$\int^{\infty}_{0}(\frac{x}{x^2+1} - \frac{C}{3x+1})dx$$

How do I do this?

The only thing I can think of is creating a common denominator?

$$\displaystyle\int\limits^{\infty}_{0} \dfrac{x^2\left(3-c\right)+x-c}{\left(x^2+1\right)\left(3x+1\right)}\,\mathrm{d}x$$

Something tells me $C = 3$ is right, but I don't know why I think that or if it's even right.

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    @Kaynex Bingo. Post that as an answer. – 2017-02-14
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    In fact, the same question was asked four hours before your question. – 2017-02-14
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    No need to downvote just cause it was answered before – 2017-02-14

1 Answers 1

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A quotient to infinity will converge if its denominator is more than one degree higher than it's numerator.

If C is anything other than 3, then the denominator is exactly one degree higher and thus the integral is not convergent. If C is exactly 3, then the denominator is two degrees higher and the integral is convergent.