Board 2023 Class 12 Maths Q5 Solution: f(x) when d/dx (f(x)) = log x

The question

If $\frac{d}{dx}(f(x)) = \log x$ ⁠, then $f(x)$ equals :

(a) $-\frac{1}{x} + C$

(b) $x(\log x - 1) + C$

(c) $x(\log x + x) + C$

(d) $\frac{1}{x} + C$

Exam answer

Since $\frac{d}{dx}(f(x)) = \log x$ ⁠,

f(x) = \int \log x\,dx

Why?

Integration undoes differentiation. If the derivative of $f(x)$ is $\log x$ ⁠, then $f(x)$ is an integral of $\log x$ ⁠, plus a constant $C$ ⁠.

Integrating by parts, with $\log x$ as the first function and 1 as the second:

Why?

$\log x$ has no basic integral of its own, but it is easy to differentiate. So write it as $\log x \cdot 1$ and use integration by parts:

\begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned}

with $u = \log x$ and $v = 1$ ⁠.

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

Why?

The pieces: $\frac{du}{dx} = \frac{1}{x}$ and $\int 1\,dx = x$ ⁠. Then $\frac{1}{x} \cdot x = 1$ ⁠, whose integral is $x$ ⁠.

\begin{aligned} u\int v\,dx &= (\log x)(x) = x\log x \\[4pt] \frac{du}{dx}\int v\,dx &= \frac{1}{x} \cdot x = 1 \\[4pt] \int 1\,dx &= x \end{aligned}

Put them into the formula: $x\log x - x + C$ ⁠. Then take out the common factor $x$ ⁠:

\begin{aligned} x\log x - x &= x\log x - x(1) \\ &= x(\log x - 1) \end{aligned}

Check by differentiating, with the product rule:

\begin{aligned} &\frac{d}{dx}\big(x\log x\big) \\ &= (1)(\log x) + (x)\left(\frac{1}{x}\right) \\ &= \log x + 1 \\[4pt] &\frac{d}{dx}\big(x\log x - x\big) \\ &= \log x + 1 - 1 = \log x \end{aligned}

Answer: (b) $x(\log x - 1) + C$

Where this comes from

Chapter 7, Integrals: Integration by parts.

Question type: Integration by parts

How to spot it: click to see

Type 8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

Here: Here $f(x) = \int\log x\,dx$ ⁠, and there is no card for $\log x$ ⁠. Write it as $\log x \cdot 1$ ⁠, with $u = \log x$ and $v = 1$ ⁠.

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q5: f(x)f(x)f(x) when ddx(f(x))=log⁡x\frac{d}{dx}(f(x)) = \log xdxd​(f(x))=logx