Question

Find or evaluate the integral using substitution first, then using integration by parts. $$ \int \ln \left(x^{2}+1\right) d x $$

Short answer

The integral evaluates to \( \frac{1}{2} (x^2 + 1) \ln(x^2 + 1) - \frac{1}{2} (x^2 + 1) + C \)

Step-by-step solution

Set up for substitution

First, we use the substitution \( u = x^2 + 1 \). The derivative \( du = 2x dx \). We can rewrite the intergral as follows \( \frac{1}{2} \int \ln(u) du \). Now the integral looks simpler.

Set up for integration by parts

Now we will use the integration by parts on the simpler representation of the integral. By using integration by parts, the formula is \(\int u v dx = u v - \int v du\). Let \( v = \ln(u) \) and \( dv = \frac{1}{u} du \). Then \( u = u \) and \( du = du \). After computing these, it is easier to compute the integral.

Apply integration by parts

Apply the formula \(\int u v dx = u v - \int v du\) as follows: \( \frac{1}{2} = \frac{1}{2} u \ln(u) - \frac{1}{2} \int du = \frac{1}{2} u \ln(u) - \frac{1}{2} u + C\). This is the result of the first method. Notice that we must convert \(u\) to \(x\).

Convert back to x

Now that we have our result in terms of \(u\), we need to change everything back in terms of \(x\). Substitute back \(u = x^2 + 1\) from Step 1 to obtain the final answer: \( \frac{1}{2} (x^2 + 1) \ln(x^2 + 1) - \frac{1}{2} (x^2 + 1) + C \).