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Question 26 of 42

Q.Evaluate : ∫1x+2−x+3 dx\int \dfrac{1}{\sqrt{x+2}-\sqrt{x+3}}\,dx

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2023Subjective· 3mImportance★★★★★
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Rationalise, then use ∫(x+a)1/2dx=23(x+a)3/2\int(x+a)^{1/2}dx=\tfrac23(x+a)^{3/2}: answer =−23[(x+2)3/2+(x+3)3/2]+c=-\tfrac23[(x+2)^{3/2}+(x+3)^{3/2}]+c.

I=∫1x+2−x+3 dx.I=\int\frac{1}{\sqrt{x+2}-\sqrt{x+3}}\,dx.

Step 1 — rationalise by multiplying numerator and denominator by the conjugate x+2+x+3\sqrt{x+2}+\sqrt{x+3}:

1x+2−x+3⋅x+2+x+3x+2+x+3=x+2+x+3(x+2)−(x+3)=x+2+x+3−1.\frac{1}{\sqrt{x+2}-\sqrt{x+3}}\cdot\frac{\sqrt{x+2}+\sqrt{x+3}}{\sqrt{x+2}+\sqrt{x+3}}=\frac{\sqrt{x+2}+\sqrt{x+3}}{(x+2)-(x+3)}=\frac{\sqrt{x+2}+\sqrt{x+3}}{-1}.

So I=−∫(x+2+x+3)dx.I=-\displaystyle\int\left(\sqrt{x+2}+\sqrt{x+3}\right)dx.

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