MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
CBSE
Class 12 Maths
Integrals Quiz 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
∫x² sin x³ dx =
0%
\(\frac{1}{3}\) cos x³ + c
0%
–\(\frac{1}{3}\) cos x + c
0%
\(\frac{-1}{3}\) cos x³ + c
0%
\(\frac{1}{2}\) sin² x³ + c
Explanation
\(\frac{-1}{3}\) cos x³ + c
∫tan √xdx is equal to
0%
(x + 1)tan √x – √x + C
0%
x tan √x – √x + C
0%
√x – x tan √x + C
0%
√x – (x + 1)tan √x + C
Explanation
(x + 1)tan √x – √x + C
∫e(\(\frac{1-x}{1+x^2}\))² dx is equal to
0%
\(\frac{e^x}{1+x^2}\) + C
0%
–\(\frac{-e^x}{1+x^2}\) + C
0%
\(\frac{e^x}{(1+x^2)^2}\) + C
0%
\(\frac{-e^x}{(1+x^2)^2}\) + C
Explanation
\(\frac{e^x}{1+x^2}\) + C
If ∫\(\frac{dx}{(x+2)(x^2+1)}\) = a log |1 + x²| + b tan x + \(\frac{1}{5}\) log |x + 2| + C, then
0%
a = \(\frac{-1}{10}\), b = \(\frac{-2}{5}\)
0%
a = \(\frac{1}{10}\), b = \(\frac{-2}{5}\)
0%
a = \(\frac{-1}{10}\), b = \(\frac{2}{5}\)
0%
a = \(\frac{1}{10}\), b = \(\frac{2}{5}\)
Explanation
a = \(\frac{-1}{10}\), b = \(\frac{2}{5}\)
∫ \(\frac{x^3}{x+1}\) is equal to
0%
x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 – x| + C
0%
x + \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 – x| + C
0%
x + \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 + x| + C
0%
x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 + x| + C
Explanation
x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 + x| + C
If ∫\(\frac{x^3dx}{\sqrt{1+x^2}}\) = a(1 + x²)+ b\(\sqrt{1 + x^2}\) + C, then
0%
a = \(\frac{1}{3}\), b = 1
0%
a = \(\frac{-1}{3}\), b = 1
0%
a = \(\frac{-1}{3}\), b = -1
0%
a = \(\frac{1}{3}\), b = -1
Explanation
a = \(\frac{1}{3}\), b = -1
Evaluate: ∫(2 tan x – 3 cot x)² dx
0%
-4tan x – cot x – 25x + C
0%
4 tan x – 9 cot x – 25x + C
0%
– 4 tan x + 9 cot x + 25x + C
0%
4 tan x + 9 cot x + 25x + C
Explanation
4 tan x – 9 cot x – 25x + C
Evaluate: ∫ sec²(7 – 4x)dx
0%
–\(\frac{1}{4}\) tan(7 – 4x) + C
0%
\(\frac{1}{4}\) tan(7 – 4x) + C
0%
\(\frac{1}{4}\) tan(7 + 4x) + C
0%
–\(\frac{1}{4}\) tan(7x – 4) + C
Explanation
–\(\frac{1}{4}\) tan(7 – 4x) + C
∫ \(\frac{10x^9+10^xlog_e 10}{10^x+x^{10}}\) dx is equal to
0%
10 – x + C
0%
10+ x+ C
0%
(10– x)+ C
0%
log (10+ x) + C
Explanation
log(10+ x) + C
Evaluate: ∫ sec x cosec xdx
0%
\(\frac{3}{5}\) tan x – 3 tan x + C
0%
–\(\frac{3}{5}\) tan x + 3 tan + C
0%
–\(\frac{3}{5}\) tan x – 3 tan + C
0%
None of these
Explanation
–\(\frac{3}{5}\) tanx + 3 tan + C
Evaluate: ∫ \(\frac{1}{\sqrt{9+8x-x^2}}\) dx
0%
-sin(\(\frac{x-4}{5}\)) + C
0%
sin(\(\frac{x+4}{5}\)) + C
0%
sin
0%
None of these
Explanation
sin(\(\frac{x-4}{5}\)) + C
Evaluate: ∫ \(\frac{1}{\sqrt{1-e^{2x}}}\) dx
0%
log |e + \(\sqrt{e^{-2x} – 1}\)| + C
0%
-log |e + \(\sqrt{e^{-2x} – 1}\)| + C
0%
-log |e– \(\sqrt{e^{-2x} – 1}\)| + C
0%
None of these
Explanation
-log |e + \(\sqrt{e^{-2x} – 1\)| + C
If ∫ \(\frac{3x+4}{x^3-2x-4}\) dx = log |x – 2| + k log f(x) + c, then
0%
f(x) = |x² + 2x + 2|
0%
f(x) = x² + 2x + 2
0%
k = –\(\frac{1}{2}\)
0%
All of these
Explanation
All of these
∫ cos(log.x)dx is equal to
0%
\(\frac{1}{2}\) x[cos (logx) + sin(log x)]
0%
x[cos (log x) + sin(log x)]
0%
\(\frac{1}{2}\) x[cos (log x) – sin(log x)]
0%
x[cos (log x) – sin(log x)]
Explanation
–\(\frac{3}{5}\) tan x + 3 tan+ C
∫ |x| dx is equal to
0%
\(\frac{1}{2}\) x² + C
0%
–\(\frac{x^2}{2}\) + C
0%
x|x| + C
0%
\(\frac{1}{2}\) x|x| + C
Explanation
\(\frac{1}{2}\) x|x| + C
∫ sin xdx is equal to
0%
cos x + C
0%
x sin x + \(\sqrt{1-x^2}\) + C
0%
\(\frac{1}{\sqrt{1-x^2}}\) + C
0%
x sin x – \(\sqrt{1-x^2}\) + C
Explanation
x sin x + \(\sqrt{1-x^2}\) + C
∫ cos (\(\frac{1}{x}\))dx equals
0%
x sec x + log |x + \(\sqrt{x^2-1}\)| + C
0%
x sec x – log |x + \(\sqrt{x^2-1}\)| + C
0%
-x sec x – log |x + \(\sqrt{x^2-1}\)| + C
0%
None of these
Explanation
x sec x – log |x + \(\sqrt{x^2-1}\)| + C
∫\(\frac{dx}{1+\cos x}\) =
0%
tan \(\frac{x}{2}\) + k
0%
\(\frac{1}{2}\) tan \(\frac{x}{2}\) + k
0%
2 tan \(\frac{x}{2}\) + k
0%
tan² \(\frac{x}{2}\) + k
Explanation
tan \(\frac{x}{2}\) + k
∫\(\frac{\cos 2x dx}{(\sin x+\cos x)^2}\) =
0%
log | sin x + cos x | + c
0%
log | sin x – cos x | + c
0%
–\(\frac{1}{\sin x+\cos x}\) + c
0%
\(\frac{1}{(\sin x+\cos x)^2}\)
Explanation
log | sin x + cos x | + c
∫\(\frac{(1+\log x)^2}{1+x^2}\) dx =
0%
\(\frac{1}{3}\)(1+log)³ + c
0%
\(\frac{1}{2}\)(1+log)² + c
0%
log (log 1 + x) + 2
0%
None of these
Explanation
\(\frac{1}{3}\)(1+log)³ + c
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)