Sample 2022-23 Class 12 Maths Q5 Solution: f(x) when f′(x) = x + 1/x

The question

If $f'(x) = x + \frac{1}{x}$ ⁠, then $f(x)$ is

(a) $x^2 + \log|x| + C$

(b) $\frac{x^2}{2} + \log|x| + C$

(c) $\frac{x}{2} + \log|x| + C$

(d) $\frac{x}{2} - \log|x| + C$

Exam answer

\begin{aligned} f(x) &= \int\left(x + \frac{1}{x}\right)dx \end{aligned}

Why?

We know the derivative $f'(x)$ and want $f(x)$ ⁠. Integration undoes differentiation, so $f(x)$ is the indefinite integral of $f'(x)$ ⁠.

\begin{aligned} f(x) &= \int x\,dx + \int\frac{1}{x}\,dx \\ &= \frac{x^2}{2} + \log|x| + C \end{aligned}

Why?

Integrate term by term with two standard results:

\begin{aligned} \int x^n\,dx &= \frac{x^{n + 1}}{n + 1} + C \\ \int\frac{1}{x}\,dx &= \log|x| + C \end{aligned}

With $n = 1$ the first gives $\frac{x^2}{2}$ ⁠. Watch out: forgetting to divide by 2 gives option (a).

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

Check by differentiating: $\frac{d}{dx}\left(\frac{x^2}{2} + \log|x|\right) = x + \frac{1}{x}$ ⁠.

\begin{aligned} \frac{d}{dx}\left(\frac{x^2}{2}\right) &= \frac{2x}{2} = x \\ \frac{d}{dx}\left(\log|x|\right) &= \frac{1}{x} \end{aligned}

Answer: (b) $\dfrac{x^2}{2} + \log|x| + C$

Where this comes from

Chapter 7, Integrals: Antiderivative from a given derivative.

Question type: Basic integrals

How to spot it: click to see

Type 1. Basic integrals: tidy up, then use the cards

Spot it: powers of $x$ ⁠, $e^x$ ⁠, sin or cos, or a fraction with a single term at the bottom.

Here: To get $f$ back from $f'(x) = x + \frac{1}{x}$ ⁠, integrate the two terms one at a time: $x$ gives $\frac{x^2}{2}$ by the power card, and $\frac{1}{x}$ gives $\log|x|$ ⁠. Then add one $+\,C$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

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