Q.Evaluate the integral using substitution $\int_{0}^{1}\frac{x}{x^2+1}\,dx$

The key idea is U Substitution: choose $u = x^2 + 1$ so that $du = 2x\,dx$, which simplifies the denominator.

Step 1: Let $u = x^2 + 1$. Then $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$.

Step 2: Change the limits: when $x = 0$, $u = 1$; when $x = 1$, $u = 2$.

Step 3: The integral becomes

Step 4: Evaluate: $\frac12 [\log u]_{1}^{2} = \frac12 (\log 2 - \log 1) = \frac12 \log 2$.

Why substitution works here

When you see a function and its derivative lurking in an integral, substitution is your best friend. Look at the denominator: $x^2+1$. Its derivative is $2x$, and the numerator has an $x$ — that’s almost the derivative, just missing a factor of 2. This is the classic signal for a u-substitution: let $u$ be the “inside” function whose derivative appears (up to a constant).

The idea is to rewrite the integral in terms of $u$, so the messy $x$ dependence disappears and we’re left with something simple like $\frac{1}{u}$ — which integrates to a natural log.

Step-by-step solution

1. Choose the substitution.

Let $u = x^2 + 1$. Then differentiate:

2. Rewrite the integrand in terms of $u$.

We have $x\,dx$ in the numerator, but $du = 2x\,dx$ gives $x\,dx = \frac{1}{2}\,du$. So the integral becomes:

3. Change the limits of integration.

Since this is a definite integral, we must update the limits for $u$:

• When $x = 0$, $u = 0^2 + 1 = 1$.
• When $x = 1$, $u = 1^2 + 1 = 2$.

So the integral becomes:

4. Integrate.

The integral of $\frac{1}{u}$ is $\log|u|$. Since $u$ is positive on $[1,2]$, we can drop the absolute value:

5. Simplify.

$\log 1 = 0$, so the result is:

Method: Substitution when the numerator is (a multiple of) the derivative of the denominator

Use this for any $\dfrac{g'(x)}{g(x)}$ pattern: the numerator is the denominator's derivative up to a constant, so the integral collapses to a logarithm.

Steps

Step 1: Check the derivative match.

Differentiate the denominator and compare with the numerator. If they agree up to a constant factor, take $u=\text{denominator}$.

Step 2: Substitute and adjust the differential.

With $u=g(x)$, $du=g'(x)\,dx$; solve for the exact group present in the integrand (e.g. $x\,dx=\tfrac12\,du$).

Step 3: Convert the limits to $u$-values.

Replace $x$-limits by $u=g(a)$ and $u=g(b)$.

Step 4: Integrate and simplify.

then evaluate between the new limits and combine logs.

Common Mistakes

Mistake 1: Not updating the limits after $u=x^2+1$.

Why it's wrong: the limits $0$ and $1$ are $x$-values; in $u$ they become $1$ and $2$. Correct approach: change the limits, or back-substitute before evaluating.

Mistake 2: Missing the $\tfrac12$ from $x\,dx=\tfrac12\,du$.

Why it's wrong: $du=2x\,dx$, so the integral is $\tfrac12\int\frac{du}{u}$, giving $\tfrac12\log 2$ not $\log 2$. Correct approach: keep the constant factor throughout.