In the print window choose “Save as PDF”. The file prints without a web address or date line.
Mere Teacher · Previous Year Questions
JEE Main Mathematics · Indefinite Integration · Set 12 (Q331-360)
Scan or visit mereteacher.com for free PYQ practice, timed CBT tests, step-by-step solutions and BroBot AI. Answers and solutions are free with a sign-up.
331.
The integral ∫(x12+3x6+1)tan−1(x3+x31)(x8−x2)dx equal to :
(1)loge((tan−1(x3+x31)))1/3+C
(2)loge((tan−1(x3+x31)))1/2+C
(3)loge((tan−1(x3+x31)))+C
(4)loge((tan−1(x3+x31)))3+C
332.
∫01/2(1+x2)1−x2dx is equal to
(1)21tan−132
(2)22tan−1(23)
(3)22tan−1(23)
(4)22tan−1(23)
333.
∫x2+2x+5dx is equal to
(1)(x+1)x2+2x+5−2log(x+1+x2+2x+5)+C
(2)21(x+1)x2+2x+5+2log(x+1+x2+2x+5)+C
(3)(x+1)x2+2x+5+2log(x+1+x2+2x+5)+C
(4)(x+1)x2+2x+5+21log(x+1+x2+2x+5)+C
334.
If k∫01x.f(3x)dx=∫03t⋅f(t)dt, then the value of k is
(1)9
(2)3
(3)1/9
(4)1/3
335.
If In=∫cosxsinnxdx, then In=
(1)−n−12cos(n−1)x−In−2
(2)n−12cos(n−1)x+In−2
(3)−n+12sin(n+1)x−In−2
(4)−n+12cos(n−1)x−In−2
336.
If ∫tan2x−cot2x1+cos8xdx=f(x)⋅cos(g(x))+c, then f(41)+g(41)=
(1)2
(2)817
(3)815
(4)1633
337.
If [x] is the greatest integer function not greater than x, then ∫011[x]dx=
(1)55
(2)45
(3)66
(4)35
338.
If f(x)=∫x1/4(1+x1/4)1dx,f(0)=−6, then f(1) is equal to:
(1)4(loge2−2)
(2)2−loge22
(3)loge2+2
(4)4loge2+2
339.
If ∫cotx−tanxcos4x+1=kcos4x+C, then k is equal to
(1)2−1
(2)8−1
(3)3−1
(4)5−1
340.
If ∫esecx(secxtanxf(x)+(secxtanx+sec2x))dx=esecxf(x)+Cthen a possible choice of f(x) is:
(1)secx+tanx+21
(2)secx−tanx−21
(3)secx+xtanx−21
(4)xsecx+tanx+21
341.
∫0π/2cot7x+tan7xtan7xdx is equal to
(1)4π
(2)2π
(3)6π
(4)3π
342.
If x2+2x+1x2+x+1=A+x+1B+(x+1)2C, then A−B is equal
(1)4C
(2)4C+1
(3)3C
(4)2C
343.
If (x2+4)(x2+9)(x2+16)2x+7=x2+4Ax+l+x2+9Bx+m+x2+16Cx+n, then A1+B1+C1=
(1)0
(2)27
(3)2105
(4)2109
344.
∫01ex+e−xdx is equal to
(1)π4-tan−1(e)
(2)tan−1(e)-π4
(3)tan−1(e)+π4
(4)tan−1(e)
345.
∫25x2+925x2+8dx=
(1)2x25x2+9+1011sinh−1(35x)+C
(2)2x25x2+9−107log(35x+25x2+9)+C
(3)2x25x2+9+107sinh−1(35x)+C
(4)2x25x2+9+1011log(35x−25x2+9)+C
346.
∑x+x−1dx=
(1)loge∣x+x−1∣−31tan−1(32x−1+1)+c
(2)31loge∣x+x−1∣−tan−1(32x−1+1)+c
(3)32loge∣x+x−1∣−tan−1(32x−1+1)+c
(4)loge∣x+x−1∣−32tan−1(32x−1+1)+c
347.
∫f(x)g(x)f(x)g’(x)−f’(x)g(x)×[logg(x)−logf(x)]dx is equal to
(1)log(f(x)g(x))+C
(2)21(logf(x)g(x))2+C
(3)f(x)g(x)log(f(x)g(x))+C
(4)log(f(x)g(x))−f(x)g(x)+C
348.
Given, (x+1)2(x+3)3x−2=x+1A+(x+1)2B+x+3C, then 4A+2B+4C is equal to
(1)5
(2)-5
(3)-3
(4)3
349.
∑x2(x2−2)5x2+3dx=
(1)2213log(2+x2−x)+2x3+C
(2)4213log(x−2x+2)+2x3+C
(3)4213log(x+2x−2)+2x3+C
(4)325log(x−2x+2)+53x+C
350.
∫(x+1)5(x3+3x2+3x+1)dx=
(1)−(x+1)1+c
(2)15log(x+1)+c
(3)log(x+1)+c
(4)tan−1x+c.
351.
If ∫x(x+1)xdx=ktan−1m, then (k,m) is
(1)(1,x)
(2)(2,x)
(3)(2,x)
(4)(1,x)
352.
If ∫8−sin2xcosx−sinxdx=asin−1(bsinx+cosx)+cwhere c is a constant of integration, then the ordered pair (a,b) is equal to :
(1)(1,−3)
(2)(1,3)
(3)(−1,3)
(4)(3,1)
353.
∫cosx(1+cosx)1−cosxdx=
(1)log|secx+tanx|−2(cosecx−cotx)+C
(2)log|secx+tanx|−2(cosecx+cotx)+C
(3)log|secx+tanx|+2(cosecx−cotx)+C
(4)log|secx+tanx|+2(cosecx+cotx)+C
354.
∫13x+4−x4−xdx=
(1)2
(2)0
(3)1
(4)3
355.
∫04π3+sin2xsinx+cosxdx is equal to
(1)21log3
(2)log2
(3)log3
(4)41log3
356.
If ∫cosx⋅cos2x⋅cos5xdx=Asin2x+Bsin4x+Csin6x+Dsin8x+k (where k is the arbitrary constant of integration), then B1+C1=
(1)A1−D1
(2)A1+D1
(3)1
(4)0
357.
If ∫2sin2x−3sinx+42sin2x−3cosxdx=f(x)+c where c is the constant of integration, then f(2π)−f(0)=
(1)2log2
(2)0
(3)log(43)
(4)1
358.
If x3+x5x2+2=xA1+x2+1A2x+A3, then (A1,A2,A3)=
(1)(0,2,3)
(2)(3,0,2)
(3)(2,3,0)
(4)(2,0,3)
359.
Let [t] denote the greatest integer function. If ∫02.4[x2]dx=α+β2+γ3+δ5, then α+β+γ+δis equal to
360.
Let fn=∫02π(∑k=1nsink−1x)(∑k=1n(2k−1)sink−1x)cosxdx,n∈N. Then f21 - f20 is equal to