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Previous year question hub

Indefinite Integration - Calculus - Mathematics Previous Year Questions

Practice Indefinite Integration - Calculus - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.

4Papers
4Years
6Questions
1Topics

Indefinite Integration question pattern

Every graph below is calculated only from this selection.

Questions by year

Year-wise coverage for Indefinite Integration. Each bar uses a separate theme-derived color.

Difficulty distribution

How the classified questions are distributed by difficulty.

Not classified 6 100%

Question type distribution

MCQ, numerical, multiple-select and other formats found in these papers.

Multiple Choices 6 100%

Subject weightage

Top subjects by unique question coverage.

Mathematics
6 Qs

Most asked topics

Top topics across the included previous year papers.

Calculus
6 Qs

Subtopic coverage

Top subtopics inside this exact selection.

Indefinite Integration
6 Qs

Paper coverage

Question coverage for the most populated papers. Every active PYP paper remains listed below.

BITSAT 2023
2 Qs
BITSAT 2022
2 Qs
BITSAT 2021
1 Qs
BITSAT 2020
1 Qs

Included previous year papers

Newest papers appear first. Sort by year, question coverage or name.

PaperYear / sessionQuestions in this viewOpen
BITSAT 202320232View paper
BITSAT 202220222View paper
BITSAT 202120211View paper
BITSAT 202020201View paper

All Indefinite Integration previous year questions

Practice every matching question in batches of 20, with every available option.

1
2020 · Mathematics · Calculus · Indefinite Integration
BITSAT 2020
\(\int {{{8{x^{43}} + 13{x^{38}}} \over {{{({x^{13}} + {x^5} + 1)}^4}}}dx}\) equals to
A
\({{{x^{39}}} \over {3{{({x^{13}} + {x^5} + 1)}^3}}} + C\)
B
\({{{x^{39}}} \over {{{({x^{13}} + {x^5} + 1)}^3}}} + C\)
C
\({{{x^{39}}} \over {5{{({x^{13}} + {x^5} + 1)}^3}}} + C\)
D
None of these
Open complete paper
2
2021 · Mathematics · Calculus · Indefinite Integration
BITSAT 2021

\(\int {{1 \over {1 - 2\sin x}}dx}\) is equal to

A
\({1 \over {2\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c\)
B
\({{\sqrt 3 } \over 2}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c\)
C
\({1 \over {\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c\)
D
None of the above
Open complete paper
3
2022 · Mathematics · Calculus · Indefinite Integration
BITSAT 2022

Let \(f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}}\) and \(f(0) = 0\), then the value of \(f(1)\) be

A
\(\log (1 + \sqrt 2 )\)
B
\(\log (1 + \sqrt 2 ) - {\pi \over 4}\)
C
\(\log (1 + \sqrt 2 ) + {\pi \over 2}\)
D
None of these
Open complete paper
4
2022 · Mathematics · Calculus · Indefinite Integration
BITSAT 2022

The value of \(\int {{1 \over {{{[{{(x - 1)}^3}{{(x + 2)}^5}]}^{{1 \over 4}}}}}dx}\), is

A
\({4 \over 3}{\left( {{{x + 1} \over {x - 2}}} \right)^{{1 \over 4}}} + C\)
B
\({3 \over 4}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C\)
C
\({4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C\)
D
\({1 \over 3}{\left( {{{2x - 1} \over {4x - 3}}} \right)^{{1 \over 4}}} + C\)
Open complete paper
5
2023 · Mathematics · Calculus · Indefinite Integration
BITSAT 2023

Let \(f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x\), where \(x \geq 0\). Then, \(f(3)-f(1)\) is equal to

A
\(\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}\)
B
\(-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}\)
C
\(-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}\)
D
\(\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}\)
Open complete paper
6
2023 · Mathematics · Calculus · Indefinite Integration
BITSAT 2023

The value of integral \(\int \frac{d x}{(1+x)^{3 / 4}(x-2)^{5 / 4}}\) is is equal to

A
\(-\frac{3}{4}\left(\frac{x+1}{x-2}\right)^{1 / 4}+C\)
B
\(-\frac{3}{4}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C\)
C
\(-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+C\)
D
\(-\frac{4}{3}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C\)
Open complete paper