# Can you evaluate #int \ (xln(x+1))/(x+1) \ dx #?

I tried this:

Have a look:

By signing up, you agree to our Terms of Service and Privacy Policy

# int \ (xln(x+1))/(x+1) \ dx = (x+1)ln(x+1) - x - 1/2 ln^2(x+1) + C #

Yes I can!

We want to find:

An obvious logical substitution would be:

Substituting into the integral gives us:

The first integral is a well known result and one that probably should be memorised. It can be derived using Integration by parts:

Then plugging into the IBP formula:

gives us:

And for the second integral we can perform another substitution:

Substituting into the second integral gives us:

Combining these two results then gives us:

By signing up, you agree to our Terms of Service and Privacy Policy

To evaluate the integral (\int \frac{x\ln(x+1)}{x+1} , dx), we can use integration by parts.

Let (u = \ln(x+1)) and (dv = \frac{x}{x+1} , dx). Then, (du = \frac{1}{x+1} , dx) and (v = \int \frac{x}{x+1} , dx).

To find (v), we can perform polynomial long division or notice that (x = (x+1) - 1), so we can rewrite the integral as (\int \left(1 - \frac{1}{x+1}\right) , dx).

Integrating term by term, we get (v = x - \ln(x+1)).

Now, we use the formula for integration by parts: (\int u , dv = uv - \int v , du).

Substituting our values in, we have:

[ \begin{aligned} \int \frac{x\ln(x+1)}{x+1} , dx &= x\ln(x+1) - \int (x - \ln(x+1)) \cdot \frac{1}{x+1} , dx \ &= x\ln(x+1) - \int \frac{x}{x+1} , dx + \int \frac{\ln(x+1)}{x+1} , dx \ &= x\ln(x+1) - (x - \ln(x+1)) + C \ &= x\ln(x+1) - x + \ln(x+1) + C, \end{aligned} ]

where (C) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7