Evaluate the indefinite integral.∫12x+5dx\int_{}^{}\frac{1}{2x+5}dx

1 2 ln ⁡ ( 2 x + 5 ) + C \frac12\ln\left(2x+5\right)+C

Evaluate the indefinite integral. ∫ 1 2 x + 5 d x \int_{}^{}\frac{1}{2x+5}dx

Here we're asked to evaluate the indefinite integral of one over two x plus five dx. Now looking at this problem, it may not immediately be clear that we need to use substitution, so it can be helpful to think of this here as one over some function raised to the first power. So this is one over two x plus five to the power of one. This might make it a little bit easier to decipher why this needs a substitution. So let's go ahead and get into our first step here which is choosing u and finding du. Now remember, u is often going to be an inside function here which is why I wrote this in parentheses raised to the first power. So if we think about this here, if I choose u to be two x plus five, that makes du equal to the derivative of that times dx, which is just going to be two dx. So when we move into step two, rewriting this integral only in terms of this u and du, this is going to give us a one over u. Now du is two dx, so that means that I need to go ahead and multiply by that missing constant two as well as its reciprocal. So multiplying by two here and then taking that reciprocal on the outside since I know that I don't need it inside of my integral. This allows me to rewrite this as one half times the integral of that two one over two x plus five. And then I'm gonna go ahead and put this two out here times two d x just so that we can clearly see we have u and we have du. So if we replace that u and du in our integral, this is one half times the integral of one over u du. So what is the integral of one over u? Moving into our step three, integrating with respect to u. Remember that the integral of one over x is the natural log of x, so this being a u does not change the fact that that is our antiderivative here. The antiderivative or our integral here is one half times the natural log of u plus that constant of integration c. So now that we have integrated here with respect to u, we could we can replace u with that original function of x, in this case two x plus five. So this gives us here a one half times a natural log of two x plus five plus that constant of integration c. And this is our final answer. Feel free to let us know if you have any questions, and I'll see you in the next one.

Evaluate the indefinite integral. ∫ 1 2 x + 5 d x \int_{}^{}\frac{1}{2x+5}dx

1 2 ln ⁡ ( 2 x + 5 ) + C \frac12\ln\left(2x+5\right)+C

1 2 ln ⁡ x + C \frac12\ln x+C

1 2 ln ⁡ ( 2 x + 5 ) 2 + C \frac12\ln\left(2x+5\right)^2+C

2 ln ⁡ ( 2 x + 5 ) + C 2\ln\left(2x+5\right)+C

8. Definite Integrals

Substitution for Indefinite Integrals

Business Calculus

8. Definite Integrals

Substitution for Indefinite Integrals

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} \frac{1}{2 x + 5} d x$

$\frac{1}{2} \ln x + C$

$\frac{1}{2} \ln {(2 x + 5)}^{2} + C$

$2 \ln (2 x + 5) + C$

$\frac{1}{2} \ln (2 x + 5) + C$

Verified step by step guidance

Step 1: Recognize that the integral ∫(1/(2x+5))dx involves a logarithmic function because the integrand is in the form of 1/(ax+b), which is a standard case for logarithmic integration.

Step 2: Use the formula for integrating functions of the form ∫(1/(ax+b))dx = (1/a)ln|ax+b| + C, where 'a' is the coefficient of x and 'b' is the constant term.

Step 3: Identify 'a' and 'b' in the given integrand. Here, 'a' = 2 and 'b' = 5. Substitute these values into the formula.

Step 4: Apply the formula: ∫(1/(2x+5))dx = (1/2)ln|2x+5| + C. This simplifies the integral into a logarithmic expression.

Step 5: Conclude that the indefinite integral evaluates to (1/2)ln(2x+5) + C, where C is the constant of integration.

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