Sample 2022-23 Class 12 Maths Q9 Solution: ∫ from 2 to 3 of x/(x² + 1) dx

The question

The value of $\int_2^3 \frac{x}{x^2 + 1}dx$ is

(a) $\log 4$

(b) $\log\frac{3}{2}$

(c) $\frac{1}{2}\log 2$

(d) $\log\frac{9}{4}$

Exam answer

Let $I = \int_2^3 \frac{x}{x^2 + 1}dx$ ⁠. Put $x^2 + 1 = t$ ⁠, $2x\,dx = dt$ ⁠.

Why?

The derivative of the denominator $x^2 + 1$ is $2x$ ⁠, and the numerator is $x$ ⁠, half of it. So substituting $t = x^2 + 1$ turns the integral into a standard one.

From $2x\,dx = dt$ we get $x\,dx = \frac{1}{2}dt$ ⁠. That is where the $\frac{1}{2}$ in front comes from.

When $x = 2$ ⁠, $t = 5$ ⁠; when $x = 3$ ⁠, $t = 10$ ⁠.

Why?

For a definite integral, change the limits to the new variable: $t = 2^2 + 1 = 5$ and $t = 3^2 + 1 = 10$ ⁠. Then there is no need to go back to $x$ ⁠.

\begin{aligned} I &= \frac{1}{2}\int_5^{10}\frac{dt}{t} = \frac{1}{2}\left[\log t\right]_5^{10} \\ &= \frac{1}{2}(\log 10 - \log 5) \\ &= \frac{1}{2}\log\frac{10}{5} \\ &= \frac{1}{2}\log 2 \end{aligned}

Why?

Replace $x^2 + 1$ by $t$ and $x\,dx$ by $\frac{1}{2}dt$ ⁠:

\begin{aligned} \frac{x\,dx}{x^2 + 1} &= \frac{\frac{1}{2}dt}{t} \\ &= \frac{1}{2} \cdot \frac{dt}{t} \end{aligned}

The standard integral is $\int\frac{dt}{t} = \log|t|$ ⁠. Here $t$ runs from 5 to 10, so $t > 0$ and $\log|t| = \log t$ ⁠.

Then the law of logarithms:

\log m - \log n = \log\frac{m}{n}

With $m = 10$ and $n = 5$ ⁠:

\begin{aligned} \log 10 - \log 5 &= \log\frac{10}{5} \\ &= \log 2 \end{aligned}

Answer: (c) $\dfrac{1}{2}\log 2$

Where this comes from

Chapter 7, Integrals: Definite integral by substitution.

Question type: Substitution + Definite integrals (combines 2 types)

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: First, the $x$ on top is half the derivative of $x^2 + 1$ at the bottom. So put $x^2 + 1 = t$ ⁠; then $x\,dx = \frac12\,dt$ ⁠.

Type 10. Definite integrals: integrate, then top minus bottom

Spot it: numbers on the integral sign. The answer is a number, with no $+\,C$ ⁠.

Here: The limits 2 and 3 become $t = 5$ and $t = 10$ ⁠. Stay in $t$ and work out $\frac12(\log 10 - \log 5) = \frac12\log 2$ ⁠, with no $+\,C$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

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