Rationalizing the denominator means rewriting a fraction so that no radical (square root, cube root, …) is left on the bottom. It is algebraic housekeeping — the fraction's value never changes, because you only ever multiply by a cleverly disguised form of 1.
Why bother? A quotient like 21 is awkward to estimate (1÷1.414), but the equal form 22 is easy (1.414÷2≈0.707). Cleaner denominators are also easier to add, compare and simplify, and most answer keys expect this final form.
Case 1 — a single square root
Multiply top and bottom by that root:
53×55=535,
because 5×5=5 is rational. In general ba=bab.
Case 2 — a sum or difference with a root
Here multiplying by the root alone fails; use the conjugate, which turns the denominator into a difference of squares:
3+72×3−73−7=32−(7)22(3−7)=22(3−7)=3−7.
For b+ca, multiply by b−cb−c; the denominator becomes b2−c, a rational number.
Watch out
Multiply both the numerator and the denominator by the same expression. Changing only the bottom changes the value of the fraction. …
Rationalizing gives 1+x+x, and ∫01(1+x+x)dx=342.
First, is it improper?
At x=0 the denominator is 1−0=1, and it stays positive across [0,1], so there is no blow-up — this is a perfectly ordinary integral. The only difficulty is cosmetic: a difference of square roots on the bottom.
Rationalize the denominator
Whenever a−b sits underneath, multiply top and bottom by the conjugate a+b, because (a−b)(a+b)=a−b:
Why it's wrong: at x=0 the denominator is 1−0=1=0, and it stays positive on [0,1]; there is no singularity. Correct approach: treat it as an ordinary integral.
Mistake 2: Not rationalising.
Why it's wrong: leaving 1+x−x1 is hard to integrate; the conjugate turns it into 1+x+x. Correct approach: multiply by the conjugate. …
The key is to simplify the integrand by rationalizing the denominator using the conjugate x−1+x, which collapses the messy fraction into a simple polynomial-like expression that integrates directly. The result is 32(1+x)3/2+C, so the correct option is (C).
The problem looks intimidating at first: a fraction with sums of square roots in both numerator and denominator. But the classic trick for expressions like x+1+x is to multiply by the conjugate x−1+x. Why? Because (a+b)(a−b)=a−b, which here becomes x−(1+x)=−1, a constant. That instantly eliminates the radicals in the denominator, leaving a much simpler expression to integrate.
Let’s work through it step by step.
Multiply numerator and denominator by the conjugate
We have
I=∫x+1+x1+x+x+x2dx.
Multiply top and bottom by x−1+x:
I=∫(x+1+x)(x−1+x)(1+x+x+x2)(x−1+x)dx.
Simplify the denominator
The denominator becomes
(x)2−(1+x)2=x−(1+x)=−1.
So the integral is
I=∫−(1+x+x+x2)(x−1+x)dx.
Expand the product
Distribute the minus sign and expand:
I=∫[−(1+x)(x−1+x)−x+x2(x−1+x)]dx.
Notice that x+x2=x(1+x)=x1+x. So the second term becomes