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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ dx (x - 1) ^ 3 / 4 (x + 2) ^ 5 / 4 = k ( x - 1 x + 2 ) ^ 1 / 4...

If ∫dx(x−1)3/4(x+2)5/4=k(x−1x+2)1/4+c∫ \dfrac{\text{dx}}{{\left(\text{x} - 1\right)}^{3 / 4} {\left(\text{x} + 2\right)}^{5 / 4}} = \text{k} {\left(\dfrac{\text{x} - 1}{\text{x} + 2}\right)}^{1 / 4} + \text{c}, then the number of divisors of 30k30k is

Correct answer

8

Step-by-step solution

I=∫dx(x−1x+2)4(x+2)2putx−1x+2=t\text{I} = ∫ \dfrac{\text{dx}}{{\left(\dfrac{\text{x} - 1}{\text{x} + 2}\right)}^{4} {\left(\text{x} + 2\right)}^{2}} \text{put} \dfrac{\text{x} - 1}{\text{x} + 2} = \text{t}

On solving, we get,

I=43(x−1x+2)14+cI = \dfrac{4}{3}\left(\dfrac{x - 1}{x + 2}\right)\dfrac{1}{4} + c

k=43⇒30k=40=23×51\text{k} = \dfrac{4}{3} ⇒ 3 0 \text{k} = 4 0 = {2}^{3} \times {5}^{1}

Hence, total divisors are 4×2=84 \times 2 = 8

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