Question
Download Solution PDFThe value of \(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
\(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\)
Concept:
Apply formula :
\(\rm \int\frac{1}{\sqrt{1+t^2}}dt =\log\left(t+\sqrt{1+t^2}\right)+C\)
Calculation:
\(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\)
\(\rm=\int \frac{x^{\frac{3}{2}}}{\sqrt{1+(x^\frac{5}{2})^2}} d x\)
Substitute \(\rm x^{\frac{5}{2}}=t\implies x^{\frac{3}{2}}dx=\frac{2}{5}dt\)
Then
\(\rm =\frac{2}{5}\int\frac{1}{\sqrt{1+t^2}}dt\)
\(\rm =\frac{2}{5}\log\left(t+\sqrt{1+t^2}\right)+C\)
Undo substitution
\(\rm =\frac{2}{5}\log\left(x^{\frac{5}{2}}+\sqrt{1+x^5}\right)+C\)
Hence option (3) is correct.
Last updated on May 26, 2025
-> AAI ATC exam date 2025 will be notified soon.
-> AAI JE ATC recruitment 2025 application form has been released at the official website. The last date to apply for AAI ATC recruitment 2025 is May 24, 2025.
-> AAI JE ATC 2025 notification is released on 4th April 2025, along with the details of application dates, eligibility, and selection process.
-> Total number of 309 vacancies are announced for the AAI JE ATC 2025 recruitment.
-> This exam is going to be conducted for the post of Junior Executive (Air Traffic Control) in Airports Authority of India (AAI).
-> The Selection of the candidates is based on the Computer Based Test, Voice Test and Test for consumption of Psychoactive Substances.
-> The AAI JE ATC Salary 2025 will be in the pay scale of Rs 40,000-3%-1,40,000 (E-1).
-> Candidates can check the AAI JE ATC Previous Year Papers to check the difficulty level of the exam.
-> Applicants can also attend the AAI JE ATC Test Series which helps in the preparation.