Skip to content
Exercise 7.2 · Q3

Q.Integrate the following function: 1x+xlog⁡x\frac{1}{x+x \log x}

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★
8% · 29/373 Questions
✓ Free question

The key idea is to factor xx from the denominator and then use the substitution u=1+log⁡xu = 1 + \log x, which simplifies the integral to ∫duu=log⁡∣u∣+C\int \frac{du}{u} = \log|u| + C. The final result is log⁡∣1+log⁡x∣+C\log|1 + \log x| + C.

We start with the integral:

∫1x+xlog⁡x dx\int \frac{1}{x + x \log x} \, dx

The denominator has a common factor of xx in both terms. Factor it out:

∫1x(1+log⁡x) dx\int \frac{1}{x(1 + \log x)} \, dx

Now, why would we think of substitution here? The expression 1+log⁡x1 + \log x appears inside the denominator, and its derivative is 1x\frac{1}{x}, which is also present in the integrand. This is the classic signal for a uu-substitution: when you see a function and its derivative (up to a constant factor) multiplied together.

Let u=1+log⁡xu = 1 + \log x. Then differentiate:

dudx=1x⇒du=1xdx\frac{du}{dx} = \frac{1}{x} \quad \Rightarrow \quad du = \frac{1}{x} dx

The integral becomes:

∫1x(1+log⁡x) dx=∫1u du\int \frac{1}{x(1 + \log x)} \, dx = \int \frac{1}{u} \, du

This is a standard integral:

∫1u du=log⁡∣u∣+C\int \frac{1}{u} \, du = \log|u| + C

Now substitute back u=1+log⁡xu = 1 + \log x:

log⁡∣1+log⁡x∣+C\log|1 + \log x| + C

Watch out

A common mistake is to forget the absolute value in the logarithm. Since log⁡x\log x is defined only for x>0x > 0, and 1+log⁡x1 + \log x could be negative for 0<x<e−10 < x < e^{-1}, the absolute value is necessary for the general antiderivative. However, if the domain is restricted to x>e−1x > e^{-1}, you can drop the absolute value.

Tip

Notice that we didn't need to expand or simplify anything beyond factoring. The substitution u=1+log⁡xu = 1 + \log x works because the derivative of log⁡x\log x is 1/x1/x, which cancels the xx in the denominator perfectly. This is a textbook example of the "function-derivative" pattern.

✓Final answer

The integral evaluates to log⁡∣1+log⁡x∣+C\boxed{\log|1 + \log x| + C}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.