# SOLVED: Recall that the formula for integration by parts is obtained from the product rule. Use similar reasoning to obtain the following integration quotient rule. Let’s assume f(x) and g(x) are diff

Indefinite Integration of a Quotient Using Substitution (Power Rule)
Indefinite Integration of a Quotient Using Substitution (Power Rule)

Get 5 free video unlocks on our app with code GOMOBILE

Snapsolve any problem by taking a picture.
Try it in the Numerade app?

Solved step-by-step

Recall that the formula for integration by parts is obtained from the product rule. Use similar reasoning to obtain the following integration quotient rule.
Let’s assume f(x) and g(x) are differentiable functions. Recall the quotient rule for differentiation:
g(x) * f'(x) – f(x) * g'(x) / (g(x))^2
Translating this equation to integral form gives the following:
âˆ« g(x) * f'(x) dx / (g(x))^2 = âˆ« f(x) * g'(x) dx / g(x)
Using the substitutions u = f(x), v = g(x), du = f'(x) dx, and dv = g'(x) dx gives:
âˆ« u dv = u * v – âˆ« v du, as was to be shown.
(D) Use the formula in part (a) to evaluate âˆ« ln(x) dx. (Use C for the constant of integration)

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

03:18

Use the quotient rule to show that $\int \frac{f^{\prime}(x)}{g(x)} d x=\frac{f(x)}{g(x)}+\int \frac{f(x) g^{\prime}(x)}{[g(x)]^{2}} d x$

03:14

(a) Integration by parts is based on what differentiation rule? Explain.(b) In your own words, state how you determine which parts of the integrand should be $u$ and $d v$

01:28

Verify the given integration result by differentiation and the Chain Rule.$$\int x e^{x} d x=x e^{x}-e^{x}+C$$

Transcript

for this problem. The artists show how to obtain integration. Quotient rule that is integral of U over V squared D U is equal to negative you over V Plus the integral of one over VD. You Now let’s assume that both F of X and G of X are differentiable functions. Now the caution rule for derivatives states that the derivative with respect to X of F of X over G of X. This is equal to G of X times F prime of x minus F of X times G prime of X all over the square of G of X. Now translating this equation to integral form gives the following, we have F of X over G of X. That’s equal to the integral of G of X times F prime of x minus F of X times G prime of X over the square of G of X. And then dx And then from here we get integral of G of X times F prime of X over the square of G of x minus f of X times G prime of X. All over the square of G of X. And then dx Now we can write this further into the integral of F prime of X over G of x minus f of X over the square of G of X times G prime of X. And then dx and then we can separate this into two integral zero. We have integral of F prime of X over G of X and then dx minus integral of F of X over G of x squared times G prime of x dx That’s the same as Integral of one over G of X times F prime of X. And then dx minus integral of F of X over the square of G of X times G prime of X. And then dx and then you…

Create an account to get free access

or

EMAIL