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Exercise 7.2 · Q36

Q.Integrate the following function: (x+1)(x+log⁡x)2x\frac{(x + 1)(x + \log x)^2}{x}

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The key idea is to rewrite the integrand so that the derivative of (x+log⁡x)(x + \log x) appears as a factor, enabling a direct uu-substitution. The integral evaluates to (x+log⁡x)33+C\frac{(x + \log x)^3}{3} + C.

Why substitution works here

When you see an expression like (x+log⁡x)2(x + \log x)^2 multiplied by something, your first instinct should be: can I find the derivative of the inside function somewhere in the integrand? Here, the inside function is u=x+log⁡xu = x + \log x, and its derivative is 1+1x1 + \frac{1}{x}. The given integrand is (x+1)(x+log⁡x)2x\frac{(x+1)(x + \log x)^2}{x}. Notice that x+1x=1+1x\frac{x+1}{x} = 1 + \frac{1}{x}, which is exactly u′u'. That’s the signal — the whole integrand is u′⋅u2u' \cdot u^2, a perfect candidate for the reverse chain rule.

Tip

Spotting u′u' multiplied by a power of uu is the most common pattern for uu-substitution. If you see (something)n(something)^n and its derivative nearby, substitution is almost always the path.

Step-by-step solution

1. Rewrite the integrand to reveal the pattern

Break the fraction into two simpler terms:

(x+1)(x+log⁡x)2x=(x+1x)(x+log⁡x)2=(1+1x)(x+log⁡x)2\frac{(x + 1)(x + \log x)^2}{x} = \left( \frac{x+1}{x} \right) (x + \log x)^2 = \left(1 + \frac{1}{x}\right) (x + \log x)^2

Now it’s clear: the factor 1+1x1 + \frac{1}{x} is the derivative of x+log⁡xx + \log x.

2. Perform the substitution

Let u=x+log⁡xu = x + \log x. Then: …

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