Start free

Mathematics · Indefinite Integration

JEE Main 2025 — 7 April, Evening Shift — Question 42

If ∫(1x+1x3)(233x−24+x−26)dx\int\left(\frac{1}{x}+\frac{1}{x^{3}}\right)\left(23 \sqrt{3 x^{-24}+x^{-26}}\right) d x =−α3(α+1)(3xβ+xγ)α+1α+C,x>0,=-\frac{\alpha}{3(\alpha+1)}\left(3 x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C, x>0,

(α,β,γ∈Z)(\alpha, \beta, \gamma \in Z), where CC is the constant of integration, then α+β+\alpha+\beta+ γ\gamma is equal to _____\_\_\_\_\_ -.

Answer: 19

Numerical answer — enter this value.

Step-by-step solution

I=∫(1x+1x3)(3x24+1x26)123dxI=\int\left(\frac{1}{x}+\frac{1}{x^{3}}\right)\left(\frac{3}{x^{24}}+\frac{1}{x^{26}}\right)^{\frac{1}{23}} d x

=∫(1x2+1x4)(3x+1x3)123dx=\int\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}\right)\left(\frac{3}{x}+\frac{1}{x^{3}}\right)^{\frac{1}{23}} d x

Put 3x+1x3=t⇒(−3x2−3x4)dx=dt\frac{3}{x}+\frac{1}{x^{3}}=t \Rightarrow\left(-\frac{3}{x^{2}}-\frac{3}{x^{4}}\right) d x=d t

⇒I=−13∫t123dt=−13t123+1123+1+C\Rightarrow \quad I=-\frac{1}{3} \int t^{\frac{1}{23}} d t=-\frac{1}{3} \frac{t^{\frac{1}{23}+1}}{\frac{1}{23}+1}+C

=−13×2324(3x−1+x−3)2423+C=-\frac{1}{3} \times \frac{23}{24}\left(3 x^{-1}+x^{-3}\right)^{\frac{24}{23}}+C

⇒α=23,β=−1,β=−3\Rightarrow \alpha=23, \beta=-1, \beta=-3

⇒α+β+γ=19\Rightarrow \alpha+\beta+\gamma=19

Answer key and solution verified before publishing.

Practise Indefinite Integration

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
7 April, Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration
Topic
Miscellaneous Types of Integrals