Problem 44 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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 \(x\).

Step 4: 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: \( \frac{1}{2} (x^2 + 1) \ln(x^2 + 1) - \frac{1}{2} (x^2 + 1) + C \).