Class 12 Integrals · Very Long Answer5 Marksindefinite_integration

Class 12 Maths · Integrals · Very Long Answer #111

Evaluate \(\int \sqrt{\frac{a + x}{a - x}}\,dx\). Work for a>0 and −a<x<a.

✅ Step-by-Step Solution

Final Answer: \(a \operatorname{asin}{\left(\frac{x}{a} \right)} - \sqrt{a^{2} - x^{2}}\) + C

Assume a>0 and −a<x<a. Because both a+x and a−x are positive, √[(a+x)/(a−x)]=(a+x)/√(a²−x²).
Split I=a∫dx/√(a²−x²)+∫x dx/√(a²−x²).
The first integral is a arcsin(x/a). For the second set u=a²−x², du=−2x dx, giving −√u.
Substitute back and combine to a arcsin(x/a)−√(a²−x²); its derivative is (a+x)/√(a²−x²), the original square root.
Therefore the required integral is \(a \operatorname{asin}{\left(\frac{x}{a} \right)} - \sqrt{a^{2} - x^{2}}\) + C.

[1 mark] Assume a>0 and −a<x<a. Because both a+x and a−x are positive, √[(a+x)/(a−x)]=(a+x)/√(a²−x²).
[1 mark] Split I=a∫dx/√(a²−x²)+∫x dx/√(a²−x²).
[1 mark] The first integral is a arcsin(x/a). For the second set u=a²−x², du=−2x dx, giving −√u.
[1 mark] Substitute back and combine to a arcsin(x/a)−√(a²−x²); its derivative is (a+x)/√(a²−x²), the original square root.
[1 mark] Therefore the required integral is \(a \operatorname{asin}{\left(\frac{x}{a} \right)} - \sqrt{a^{2} - x^{2}}\) + C.

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