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Exercise 11.5 · Q16

Q.1x+3−x−4\dfrac{1}{\sqrt{x+3}-\sqrt{x-4}}

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A denominator that is a difference of two square roots is standard practice to rationalise using the conjugate.

Step 1. Multiply by the conjugate. 1x+3−x−4⋅x+3+x−4x+3+x−4=x+3+x−4(x+3)−(x−4)=x+3+x−47\dfrac1{\sqrt{x+3}-\sqrt{x-4}}\cdot\dfrac{\sqrt{x+3}+\sqrt{x-4}}{\sqrt{x+3}+\sqrt{x-4}}=\dfrac{\sqrt{x+3}+\sqrt{x-4}}{(x+3)-(x-4)}=\dfrac{\sqrt{x+3}+\sqrt{x-4}}{7}.

Step 2. Integrate each power. ∫(x+3)1/27dx=17⋅23(x+3)3/2=221(x+3)3/2\displaystyle\int\dfrac{(x+3)^{1/2}}7dx=\dfrac17\cdot\dfrac23(x+3)^{3/2}=\dfrac2{21}(x+3)^{3/2}, and similarly ∫(x−4)1/27dx=221(x−4)3/2\displaystyle\int\dfrac{(x-4)^{1/2}}7dx=\dfrac2{21}(x-4)^{3/2}. …

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