Q.$\int_{0}^{1} x\log(1+2x)\,dx$

The key idea is to integrate using integration by parts, where the logarithmic term is set as the first function.

Let $I = \int_{0}^{1} x \log(1+2x) \, dx$.

Step 1: Apply integration by parts: $\int u \, dv = uv - \int v \, du$.

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

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

Step 2:

Step 3: Simplify the rational integrand by polynomial division:

$\frac{x^2}{1+2x} = \frac{x}{2} - \frac{1}{4} + \frac{1}{4(1+2x)}$.

Thus, …

1. Integration by parts. Take $u=\log(1+2x)$, $dv=x\,dx$, so $du=\dfrac{2}{1+2x}\,dx$ and $v=\dfrac{x^2}{2}$:

The boundary term is $\tfrac12\log 3 - 0 = \tfrac12\log 3$.

2. Reduce the rational integral. Polynomial division gives

Then …

Method: Integration by parts with a logarithm (ILATE)

For $\int(\text{polynomial})\log(\cdots)\,dx$, take the logarithm as the first function so differentiating removes it.

Steps

Step 1: Choose $u$ and $dv$ by ILATE.

Logarithm beats algebraic, so $u=\log(\cdots)$ and $dv=(\text{polynomial})\,dx$. Then $du$ is a rational function and $v$ is a polynomial.

Step 2: Apply $\int u\,dv=uv-\int v\,du$. …

Common Mistakes

Mistake 1: Choosing $u=x$ instead of $u=\log(1+2x)$.

Why it's wrong: by ILATE the logarithm is the first function; picking the polynomial leaves a harder $\int x^2\cdot(\log)'$ term. Correct approach: set $u=\log(1+2x)$, $dv=x\,dx$.

Mistake 2: Forgetting the chain-rule factor in $\frac{d}{dx}\log(1+2x)$. …