Integrate: 3x^2/(x^6+1).
Hint. Put t=x^3, so dt=3x^2 dx, and note x^6=(x^3)^2 turns the denominator into t^2+1.
With t=x^3, dt=3x^2dx exactly matches the numerator, and x^6+1=t^2+1, so the integral becomes integral of dt/(t^2+1) = tan^-1(t).
✦ tan^-1(x^3) + C
Integrate: 1/(1+4x^2).
Hint. Put u=2x, so du=2dx, reducing this to the standard form 1/(1+u^2).
With u=2x, du=2dx, the integral becomes (1/2) integral of du/(1+u^2) = (1/2)tan^-1(u), giving the answer once u is replaced back by 2x.
✦ (1/2) tan^-1(2x) + C
Integrate: 1/[(2-x)^2+1].
Hint. Put t=x-2 (equivalently 2-x, since it is squared), reducing this to the standard form 1/(t^2+1).
Since (2-x)^2=(x-2)^2, putting t=x-2 gives dt=dx, and the integral becomes integral of dt/(t^2+1) = tan^-1(t).
✦ tan^-1(x-2) + C
Integrate: 1/sqrt(9-25x^2).
Hint. Factor 9-25x^2 = 25(9/25 - x^2), matching the standard form 1/sqrt(a^2-x^2) with a=3/5.
Writing 9-25x^2 = 25[(3/5)^2-x^2], the square root pulls out a factor of 5, giving (1/5) integral of dx/sqrt((3/5)^2-x^2) = (1/5)sin^-1(x/(3/5)) = (1/5)sin^-1(5x/3).
✦ (1/5) sin^-1(5x/3) + C
Integrate: 3x/(1+2x^4).
Hint. Put t=x^2, so dt=2x dx, reducing the quartic denominator to a quadratic in t.
With t=x^2, dt=2xdx, so 3x dx = (3/2)dt, and 1+2x^4=1+2t^2, giving (3/2) integral of dt/(1+2t^2), a standard tan^-1 form after factoring out the 2.
✦ (3 sqrt(2)/4) tan^-1(sqrt(2) x^2) + C
Integrate: x^2/(1-x^6).
Hint. Put t=x^3, so dt=3x^2 dx, and 1-x^6=1-t^2.
With t=x^3, dt=3x^2dx, the integral becomes (1/3) integral of dt/(1-t^2), a standard log form since the denominator factors as (1-t)(1+t).
✦ (1/6) log|(1+x^3)/(1-x^3)| + C
Integrate: (x-1)/sqrt(x^2-1).
Hint. Split into x/sqrt(x^2-1) minus 1/sqrt(x^2-1), and handle each piece with a different standard technique.
The first piece, x/sqrt(x^2-1), integrates via u=x^2-1 to sqrt(x^2-1). The second piece, 1/sqrt(x^2-1), is the standard log form log|x+sqrt(x^2-1)|. Subtracting gives the result.
✦ sqrt(x^2-1) - log|x+sqrt(x^2-1)| + C
Integrate: x^2/sqrt(x^6+a^6).
Hint. Put t=x^3, so dt=3x^2 dx, and note a^6=(a^3)^2, matching the standard form 1/sqrt(t^2+A^2).
Putting t=x^3 gives dt=3x^2dx, so the integral becomes (1/3) integral of dt/sqrt(t^2+a^6), the standard log form with A=a^3, giving (1/3)log|t+sqrt(t^2+a^6)|.
✦ (1/3) log|x^3 + sqrt(x^6+a^6)| + C
Integrate: sec^2 x/sqrt(tan^2 x+4).
Hint. Put t=tan x, so dt=sec^2 x dx — exactly the numerator.
With t=tanx, dt=sec^2xdx, the integral becomes integral of dt/sqrt(t^2+4), the standard log form, giving log|t+sqrt(t^2+4)|.
✦ log|tan x + sqrt(tan^2 x+4)| + C
Integrate: 1/sqrt(x^2+2x+2).
Hint. Complete the square: x^2+2x+2=(x+1)^2+1, matching the standard form 1/sqrt(u^2+a^2).
Completing the square gives (x+1)^2+1, so with u=x+1, the integral matches the standard form log|u+sqrt(u^2+1)|.
✦ log|x+1+sqrt(x^2+2x+2)| + C
Integrate: 1/sqrt(9x^2+6x+5).
Hint. Complete the square inside the root, then match the standard log form.
Completing the square, 9x^2+6x+5 = 9(x+1/3)^2+4, so the integral matches the standard form (1/3)log|u+sqrt(u^2+4/9)| with u=x+1/3, which simplifies using log properties.
✦ (1/3) log|3x+1+sqrt(9x^2+6x+5)| + C
Integrate: 1/sqrt(7-6x-x^2).
Hint. Complete the square: 7-6x-x^2 = 16-(x+3)^2, matching the standard form 1/sqrt(a^2-u^2).
Completing the square, 7-6x-x^2 = -(x^2+6x-7) = -(x+3)^2+16, so the integral matches the standard form sin^-1(u/4) with u=x+3.
✦ sin^-1((x+3)/4) + C
Integrate: 1/sqrt[(x-1)(x-2)].
Hint. Expand the product and complete the square, matching the standard log form.
Expanding, (x-1)(x-2)=x^2-3x+2=(x-3/2)^2-1/4, so the integral matches the standard form log|u+sqrt(u^2-1/4)| with u=x-3/2.
✦ log|x - 3/2 + sqrt(x^2-3x+2)| + C
Integrate: 1/sqrt(8+3x-x^2).
Hint. Complete the square: 8+3x-x^2 = 41/4-(x-3/2)^2, matching the standard form 1/sqrt(a^2-u^2).
Completing the square, 8+3x-x^2 = -(x-3/2)^2+41/4, so the integral matches the standard form sin^-1(u/a) with u=x-3/2, a=sqrt(41)/2.
✦ sin^-1((2x-3)/sqrt(41)) + C
Integrate: 1/sqrt[(x-a)(x-b)].
Hint. Expand the product and complete the square in terms of the midpoint (a+b)/2.
Expanding, (x-a)(x-b) = [x-(a+b)/2]^2 - [(a-b)/2]^2, so the integral matches the standard log form with u=x-(a+b)/2.
✦ log|x - (a+b)/2 + sqrt((x-a)(x-b))| + C
Integrate: (4x+1)/sqrt(2x^2+x-3).
Hint. Check whether the numerator is exactly the derivative of the expression under the root — it is.
Since d/dx(2x^2+x-3)=4x+1, exactly the numerator, put t=2x^2+x-3 so dt=(4x+1)dx, giving integral of t^-1/2 dt = 2 sqrt(t).
✦ 2 sqrt(2x^2+x-3) + C
Integrate: (x+2)/sqrt(x^2-1).
Hint. Split into x/sqrt(x^2-1) plus 2/sqrt(x^2-1), and handle each with a different standard technique.
The first piece integrates via u=x^2-1 to sqrt(x^2-1). The second piece is the standard log form 2log|x+sqrt(x^2-1)|. Adding gives the result.
✦ sqrt(x^2-1) + 2 log|x+sqrt(x^2-1)| + C
Integrate: (5x-2)/(1+2x+3x^2).
Hint. Write the numerator as a multiple of the denominator's derivative (6x+2) plus a constant remainder, then split into two integrals.
Since the denominator's derivative is 6x+2, writing 5x-2=(5/6)(6x+2)-11/3 splits the integral into (5/6) times the log of the denominator, plus a standard tan^-1 form after completing the square on 3x^2+2x+1.
✦ (5/6) log|3x^2+2x+1| - (11 sqrt(2)/6) tan^-1((3x+1)/sqrt(2)) + C
Integrate: (6x+7)/sqrt[(x-5)(x-4)].
Hint. Write the numerator as a multiple of the derivative of x^2-9x+20 plus a constant remainder, then split into two integrals.
Because the derivative of x^2-9x+20 is 2x-9, writing 6x+7=3(2x-9)+34 splits the integral into a square-root term from the first piece, plus a standard log form from the second piece after completing the square.
✦ 6 sqrt[(x-5)(x-4)] + 34 log|x - 9/2 + sqrt[(x-5)(x-4)]| + C
Integrate: (x+2)/sqrt(4x-x^2).
Hint. Write the numerator as a multiple of the derivative of 4x-x^2 plus a constant remainder, then split into two integrals.
The derivative of 4x-x^2 is 4-2x, so writing x+2=-(1/2)(4-2x)+4 splits the integral into a square-root term from the first piece, plus a standard sin^-1 form from the second piece after completing the square.
✦ -sqrt(4x-x^2) + 4 sin^-1((x-2)/2) + C
Integrate: (x+2)/sqrt(x^2+2x+3).
Hint. Write the numerator as a multiple of the derivative of x^2+2x+3 plus a constant remainder, then split into two integrals.
Since d/dx(x^2+2x+3)=2x+2, writing x+2=(1/2)(2x+2)+1 splits the integral into a square-root term from the first piece, plus a standard log form from the second piece after completing the square.
✦ sqrt(x^2+2x+3) + log|x+1+sqrt(x^2+2x+3)| + C
Integrate: (x+3)/(x^2-2x-5).
Hint. Write the numerator as a multiple of the derivative of x^2-2x-5 plus a constant remainder, then split into two integrals.
Since the derivative of x^2-2x-5 is 2x-2, x+3 can be written as (1/2)(2x-2)+4, splitting the integral into a log term from the first piece, plus a standard log-ratio form from the second piece.
✦ (1/2) log|x^2-2x-5| + (2/sqrt(6)) log|(x-1-sqrt(6))/(x-1+sqrt(6))| + C
Integrate: (5x+3)/sqrt(x^2+4x+10).
Hint. Write the numerator as a multiple of the derivative of x^2+4x+10 plus a constant remainder, then split into two integrals.
Because d/dx(x^2+4x+10)=2x+4, writing 5x+3=(5/2)(2x+4)-7 splits the integral into a square-root term from the first piece, plus a standard log form from the second piece after completing the square.
✦ 5 sqrt(x^2+4x+10) - 7 log|x+2+sqrt(x^2+4x+10)| + C
Choose the correct answer: integral of dx/(x^2+2x+2) equals (A) x tan^-1(x+1)+C (B) tan^-1(x+1)+C (C) (x+1)tan^-1x+C (D) tan^-1x+C
Hint. Complete the square in the denominator first: x^2+2x+2=(x+1)^2+1.
Completing the square gives (x+1)^2+1, so the integral matches the standard form tan^-1(u) with u=x+1.
✦ (B) tan^-1(x+1) + C
Choose the correct answer: integral of dx/sqrt(9x-4x^2) equals (A) (1/9)sin^-1((9x-8)/8)+C (B) (1/2)sin^-1((8x-9)/9)+C (C) (1/3)sin^-1((9x-8)/8)+C (D) (1/2)sin^-1((9x-8)/9)+C
Hint. Complete the square: 9x-4x^2 = -4(x-9/8)^2+81/16, matching the standard sin^-1 form.
Completing the square, 9x-4x^2 = 81/16 - 4(x-9/8)^2, so sqrt(9x-4x^2) = 2sqrt(81/64-(x-9/8)^2), giving the standard form (1/2)sin^-1[(x-9/8)/(9/8)] = (1/2)sin^-1((8x-9)/9).
✦ (B) (1/2) sin^-1((8x-9)/9) + C
Solutions written by the tuition.in editorial team and checked against lemh201.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.
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