Q.Integrate the following function: $(x^2+1) \log x$

The key idea is integration by parts, where we treat $\log x$ as the first function (since its derivative is simpler) and $x^2+1$ as the second function.

Let $I = \int (x^2+1) \log x \, dx$.

Using integration by parts: $\int u \, dv = uv - \int v \, du$, take $u = \log x$ and $dv = (x^2+1)\,dx$.

Then $du = \frac{1}{x} dx$ and $v = \int (x^2+1)\,dx = \frac{x^3}{3} + x$.

So

$I = \left(\frac{x^3}{3} + x\right) \log x - \int \left(\frac{x^3}{3} + x\right) \cdot \frac{1}{x} \, dx = \left(\frac{x^3}{3} + x\right) \log x - \int \left(\frac{x^2}{3} + 1\right) dx.$ …

We start with the integral:

The function is a product of a polynomial $(x^2+1)$ and $\log x$. There is no direct product rule in integration, but we can use integration by parts — which is the reverse of the product rule for differentiation. The formula is:

Here, we choose $u = \log x$ because its derivative $\frac{1}{x}$ is simpler, and $dv = (x^2+1) \, dx$ because its integral is a polynomial.

Let’s work through it step by step.

1. Set up integration by parts

   Let $u = \log x$ and $dv = (x^2 + 1) \, dx$.

   Then $du = \frac{1}{x} \, dx$, and $v = \int (x^2 + 1) \, dx = \frac{x^3}{3} + x$.

2. Apply the formula

3. Simplify the new integral Inside the integral:

So we have:

4. Integrate the polynomial $\int \left( \frac{x^2}{3} + 1 \right) dx = \frac{1}{3} \cdot \frac{x^3}{3} + x = \frac{x^3}{9} + x$ …

Method: Split, then integrate by parts term by term

When a polynomial multiplies a logarithm, e.g. $(x^{2}+1)\log x$, distribute the log over the polynomial and integrate each $x^{k}\log x$ piece by parts.

Steps

Step 1: Distribute.

You can also keep it as one integral with $dv=(x^{2}+1)\,dx$, $v=\dfrac{x^{3}}{3}+x$ — either is valid.

Step 2: For each piece, log is $u$ (ILATE).

Step 3: Apply by parts to each term. …

Common Mistakes

Mistake 1: Distributing the polynomial but forgetting $\int\log x\,dx$ is itself a by-parts case.

Why it's wrong: $\int\log x\,dx=x\log x-x$, not $\log x\cdot x$. Correct approach: integrate the lone $\log x$ by parts too.

Mistake 2: Writing two separate constants.

Why it's wrong: the whole expression needs only one arbitrary constant. Correct approach: combine into a single $+C$. …