Calculate the integral: integral from 5 to 0 dx/(2x+

Calculate the integral: $\int_5^0 \frac{dx}{2x+\sqrt{3x+1}}$.

We will use substitution to simplify this integral.

Step 1: Note the order of limits

The limits are reversed, so we can reverse them:

\[\int_5^0 \frac{dx}{2x+\sqrt{3x+1}} = -\int_0^5 \frac{dx}{2x+\sqrt{3x+1}}\]

We'll evaluate the integral from 0 to 5, then negate the result.

Step 2: Use substitution

Let $u = \sqrt{3x+1}$. Then $u^2 = 3x+1$, so $x = \frac{u^2-1}{3}$ and $dx = \frac{2u}{3} \, du$.

Step 3: Change the limits of integration

When $x = 0$: $u = 1$. When $x = 5$: $u = \sqrt{16} = 4$.

Step 4: Express the denominator in terms of $u$

\[2x + \sqrt{3x+1} = \frac{2(u^2-1)}{3} + u = \frac{2u^2+3u-2}{3}\]

Step 5: Rewrite the integral

\[\int_0^5 \frac{dx}{2x+\sqrt{3x+1}} = \int_1^4 \frac{3}{2u^2+3u-2} \cdot \frac{2u}{3} \, du = \int_1^4 \frac{2u}{2u^2+3u-2} \, du\]

Step 6: Factor the denominator

\[2u^2 + 3u - 2 = (2u - 1)(u + 2)\]

Step 7: Use partial fractions

\[\frac{2u}{(2u-1)(u+2)} = \frac{A}{2u-1} + \frac{B}{u+2}\]

Solving: $2u = A(u+2) + B(2u-1)$. Setting $u=\frac12$: $A=\frac25$. Setting $u=-2$: $B=\frac45$.

\[\frac{2u}{(2u-1)(u+2)} = \frac{2}{5(2u-1)} + \frac{4}{5(u+2)}\]

Step 8: Integrate

\[\int_1^4 \frac{2u}{2u^2+3u-2} \, du = \frac{2}{5}\int_1^4 \frac{du}{2u-1} + \frac{4}{5}\int_1^4 \frac{du}{u+2}\]

For the first, $\int \frac{du}{2u-1} = \frac{1}{2}\ln|2u-1|$; for the second, $\int \frac{du}{u+2} = \ln|u+2|$. Therefore:

\[= \frac{1}{5}\left[\ln|2u-1|\right]_1^4 + \frac{4}{5}\left[\ln|u+2|\right]_1^4\]

Step 9: Evaluate at the limits

\[= \frac{1}{5}[\ln(7) - \ln(1)] + \frac{4}{5}[\ln(6) - \ln(3)] = \frac{1}{5}\ln(7) + \frac{4}{5}\ln(2) = \frac{1}{5}[\ln(7) + \ln(2^4)] = \frac{1}{5}\ln(112)\]

Step 10: Apply the sign from reversed limits

\[\int_5^0 \frac{dx}{2x+\sqrt{3x+1}} = -\frac{1}{5}\ln(112)\]

Numerical value:

Using $\ln(112) \approx 4.7185$: $-\frac{1}{5}\ln(112) \approx -0.9437$.

Final answer:

\[\boxed{\int_5^0 \frac{dx}{2x+\sqrt{3x+1}} = -\frac{1}{5}\ln(112) = -\frac{\ln(112)}{5}}\]

Or equivalently:

\[\boxed{\int_5^0 \frac{dx}{2x+\sqrt{3x+1}} = -\frac{1}{5}(\ln 7 + 4\ln 2)}\]