Find the value of the constant C for which the integral ∫0∞(xx2+1−C3x+1)dx converges. Evaluate the integral for this value of C.

Step 1∫0∞(xx2+1−C3x+1)&nbsp;dx&nbsp;=limt→∞∫0t(xx2+1−C3x+1)&nbsp;dx&nbsp;Integrate the integral on the right first.∫0t(xx2+1−C3x+1)&nbsp;dx&nbsp;=(12ln|x2+1|−C3ln|3x+1|)∣0t=16(ln⁡(x2+1)3−ln⁡(3x+1)2C)∣0t=16ln⁡(x2+1)3(3x+1)2C∣0t=16ln⁡(t2+1)3(3t+1)2C−16ln⁡11=16ln⁡(t2+1)3(3t+1)2CTherefore∫0∞(xx2+1−C3x+1)&nbsp;dx&nbsp;=limt→∞(16ln⁡(t2+1)(3t+1)2C)=16ln⁡(limt→∞(t2+1)3(3t+1)2C)&nbsp;

Find out the answer to "Find the value of the constant C for..."

Ramsey

Answered question

2021-10-23

Find the value of the constant C for which the integral

\int_{0}^{\infty} (\frac{x}{x_{2} + 1} - \frac{C}{3 x + 1}) d x

converges. Evaluate the integral for this value of C.

Answer & Explanation

Usamah Prosser

Skilled2021-10-24Added 86 answers

Step 1
$\int_{0}^{\infty} (\frac{x}{x_{2} + 1} - \frac{C}{3 x + 1}) d x = lim_{t \to \infty} \int_{0}^{t} (\frac{x}{x^{2} + 1} - \frac{C}{3 x + 1}) d x$
Integrate the integral on the right first.
$\int_{0}^{t} (\frac{x}{x^{2} + 1} - \frac{C}{3 x + 1}) d x = (\frac{1}{2} \ln | x^{2} + 1 | - \frac{C}{3} \ln | 3 x + 1 |) ∣_{0}^{t}$
$= \frac{1}{6} ({\ln (x^{2} + 1)}^{3} - {\ln (3 x + 1)}^{2 C}) ∣_{0}^{t}$
$= \frac{1}{6} \ln \frac{{(x^{2} + 1)}^{3}}{{(3 x + 1)}^{2 C}} ∣_{0}^{t}$
$= \frac{1}{6} \ln \frac{{(t^{2} + 1)}^{3}}{{(3 t + 1)}^{2 C}} - \frac{1}{6} \ln \frac{1}{1}$
$= \frac{1}{6} \ln \frac{{(t^{2} + 1)}^{3}}{{(3 t + 1)}^{2 C}}$
Therefore
$\int_{0}^{\infty} (\frac{x}{x^{2} + 1} - \frac{C}{3 x + 1}) d x = lim_{t \to \infty} (\frac{1}{6} \ln \frac{(t^{2} + 1)}{{(3 t + 1)}^{2 C}})$
$= \frac{1}{6} \ln (lim_{t \to \infty} \frac{{(t^{2} + 1)}^{3}}{{(3 t + 1)}^{2 C}})$

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