Integrate: x/[(x+1)(x+2)].
Hint. Decompose into A/(x+1)+B/(x+2), and find A, B by clearing denominators and substituting the roots.
Writing x=A(x+2)+B(x+1), substituting x=-1 gives A=-1, and x=-2 gives B=2. So the fraction splits into -1/(x+1)+2/(x+2), and each piece integrates to a log.
✦ -log|x+1| + 2log|x+2| + C
Integrate: 1/(x^2-9).
Hint. Factor as (x-3)(x+3) and apply the standard partial-fraction form for 1/(x^2-a^2).
Since x^2-9=(x-3)(x+3), this matches the standard form 1/(x^2-a^2) with a=3, giving the log-ratio antiderivative directly.
✦ (1/6) log|(x-3)/(x+3)| + C
Integrate: (3x-1)/[(x-1)(x-2)(x-3)].
Hint. Decompose into A/(x-1)+B/(x-2)+C/(x-3), finding each constant by substituting the corresponding root.
Clearing denominators and substituting x=1, x=2, x=3 in turn gives A=1, B=-5, C=4, so the fraction splits into three simple log terms.
✦ log|x-1| - 5log|x-2| + 4log|x-3| + C
Integrate: x/[(x-1)(x-2)(x-3)].
Hint. Decompose into A/(x-1)+B/(x-2)+C/(x-3), finding each constant by substituting the corresponding root.
Substituting x=1, x=2, x=3 in turn (after clearing denominators) gives A=1/2, B=-2, C=3/2, so the fraction splits into three simple log terms.
✦ (1/2) log|x-1| - 2log|x-2| + (3/2)log|x-3| + C
Integrate: 2x/(x^2+3x+2).
Hint. Factor the denominator as (x+1)(x+2), then decompose into partial fractions.
Since x^2+3x+2=(x+1)(x+2), writing 2x=A(x+2)+B(x+1) and substituting the roots gives A=-2, B=4, so the integral is a combination of two logs.
✦ -2log|x+1| + 4log|x+2| + C
Integrate: (1-x^2)/[x(1-2x)].
Hint. This fraction is improper (numerator and denominator both degree 2) — divide first, then decompose the proper remainder into partial fractions.
Dividing out the improper part leaves a constant 1/2 plus a proper fraction (1-x/2)/[x(1-2x)], which decomposes into 1/x + (3/2)/(1-2x); integrating each piece gives the result.
✦ x/2 + log|x| - (3/4) log|1-2x| + C
Integrate: x/[(x^2+1)(x-1)].
Hint. Decompose into A/(x-1) + (Bx+C)/(x^2+1), finding A by substituting x=1 and B, C by matching coefficients.
Substituting x=1 gives A=1/2; matching the x^2 and constant coefficients then gives B=-1/2 and C=1/2. Integrating the linear-over-quadratic piece splits further into a log and a tan^-1 term.
✦ (1/2) log|x-1| - (1/4) log(x^2+1) + (1/2) tan^-1 x + C
Integrate: x/[(x-1)^2(x+2)].
Hint. Decompose into A/(x-1)+B/(x-1)^2+C/(x+2), since (x-1) is a repeated factor.
Clearing denominators and substituting x=1 gives B=1/3, x=-2 gives C=-2/9, and matching the x^2 coefficient gives A=2/9. The B/(x-1)^2 term integrates to -B/(x-1).
✦ (2/9) log|x-1| - 1/(3(x-1)) - (2/9) log|x+2| + C
Integrate: (3x+5)/(x^3-x^2-x+1).
Hint. Factor the cubic denominator by grouping first — it has a repeated linear factor.
Factoring by grouping, x^3-x^2-x+1=(x-1)^2(x+1), so decomposing into A/(x-1)+B/(x-1)^2+C/(x+1) and substituting the roots gives A=-1/2, B=4, C=1/2.
✦ -(1/2) log|x-1| - 4/(x-1) + (1/2) log|x+1| + C
Integrate: (2x-3)/[(x^2-1)(2x+3)].
Hint. Factor x^2-1 as (x-1)(x+1) and decompose into three simple linear-denominator fractions.
Decomposing into A/(x-1)+B/(x+1)+C/(2x+3) and substituting the three roots gives A=-1/10, B=5/2, C=-24/5, so the integral is a combination of three logs.
✦ -(1/10) log|x-1| + (5/2) log|x+1| - (12/5) log|2x+3| + C
Integrate: 5x/[(x+1)(x^2-4)].
Hint. Factor x^2-4 as (x-2)(x+2) and decompose into three simple linear-denominator fractions.
Decomposing into A/(x+1)+B/(x-2)+C/(x+2) and substituting the three roots gives A=5/3, B=5/6, C=-5/2, so the integral is a combination of three logs.
✦ (5/3) log|x+1| + (5/6) log|x-2| - (5/2) log|x+2| + C
Integrate: (x^3+x+1)/(x^2-1).
Hint. This fraction is improper (degree 3 over degree 2) — divide first, leaving a linear quotient plus a proper remainder.
Dividing, x^3+x+1 = x(x^2-1)+2x+1, so the fraction equals x + (2x+1)/[(x-1)(x+1)]. Decomposing the remainder into partial fractions with A=3/2, B=1/2 completes the integral.
✦ x^2/2 + (3/2) log|x-1| + (1/2) log|x+1| + C
Integrate: 2/[(1-x)(1+x^2)].
Hint. Decompose into A/(1-x) + (Bx+C)/(1+x^2), finding A by substituting x=1 and B, C by matching coefficients.
Substituting x=1 gives A=1, and matching the x^2 and constant coefficients gives B=1, C=1. Integrating the linear-over-quadratic piece splits into a log and a tan^-1 term.
✦ -log|1-x| + (1/2) log(1+x^2) + tan^-1 x + C
Integrate: (3x-1)/(x+2)^2.
Hint. Substitute t=x+2 to rewrite the numerator in terms of t, splitting the fraction into a 1/t piece and a 1/t^2 piece.
With t=x+2, x=t-2, the numerator becomes 3t-7, so the fraction splits into 3/t - 7/t^2, and integrating each power of t gives the result.
✦ 3 log|x+2| + 7/(x+2) + C
Integrate: 1/(x^4-1).
Hint. Factor completely as (x-1)(x+1)(x^2+1), then decompose into partial fractions with one quadratic-denominator term.
Decomposing into A/(x-1)+B/(x+1)+(Cx+D)/(x^2+1) and matching coefficients gives A=1/4, B=-1/4, C=0, D=-1/2, so the quadratic term contributes a pure tan^-1 piece.
✦ (1/4) log|(x-1)/(x+1)| - (1/2) tan^-1 x + C
Integrate: 1/[x(x^n+1)]. [Hint: multiply numerator and denominator by x^(n-1), put x^n=t]
Hint. Following the hint, multiplying top and bottom by x^(n-1) turns this into a function of t=x^n alone.
Multiplying by x^(n-1)/x^(n-1) gives x^(n-1)/[x^n(x^n+1)]. With t=x^n, dt=n x^(n-1)dx, the integral becomes (1/n) integral of dt/[t(t+1)], which decomposes into 1/t - 1/(t+1).
✦ (1/n) log|x^n / (x^n+1)| + C
Integrate: cos x/[(1-sin x)(2-sin x)]. [Hint: put sin x=t]
Hint. Following the hint, put t=sin x so dt=cos x dx, turning this into a rational function of t.
With t=sinx, dt=cosxdx, the integral becomes integral of dt/[(1-t)(2-t)], which decomposes into 1/(1-t) - 1/(2-t), each integrating to a log.
✦ log|(2 - sin x)/(1 - sin x)| + C
Integrate: (x^2+1)(x^2+2)/[(x^2+3)(x^2+4)].
Hint. Treat x^2 as a single variable y — this becomes an improper rational function of y that needs division first.
Since y=x^2 makes (y+1)(y+2)/[(y+3)(y+4)] an improper fraction (same degree top and bottom), dividing out leaves 1 plus a proper fraction that decomposes into 2/(x^2+3) - 6/(x^2+4), each a standard tan^-1 form.
✦ x + (2/sqrt(3)) tan^-1(x/sqrt(3)) - 3 tan^-1(x/2) + C
Integrate: 2x/[(x^2+1)(x^2+3)].
Hint. Put t=x^2 so dt=2x dx, reducing this to a partial-fraction problem in t alone.
With t=x^2, dt=2xdx exactly matches the numerator, so the integral becomes integral of dt/[(t+1)(t+3)], which decomposes into (1/2)/(t+1) - (1/2)/(t+3).
✦ (1/2) log[(x^2+1)/(x^2+3)] + C
Integrate: 1/[x(x^4-1)].
Hint. Factor x^4-1 completely as x(x-1)(x+1)(x^2+1) is not quite right — factor the quartic as (x-1)(x+1)(x^2+1) and keep the extra x separately, then decompose into four partial-fraction terms.
Decomposing 1/[x(x-1)(x+1)(x^2+1)] into A/x+B/(x-1)+C/(x+1)+(Dx+E)/(x^2+1) and matching coefficients gives A=-1, B=1/4, C=1/4, D=1/2, E=0.
✦ -log|x| + (1/4) log|x-1| + (1/4) log|x+1| + (1/4) log(x^2+1) + C
Integrate: 1/(e^x-1). [Hint: put e^x=t]
Hint. Following the hint, put t=e^x so dx=dt/t, turning this into a rational function of t.
With t=e^x, dt=e^xdx=tdx, so dx=dt/t, and the integral becomes integral of dt/[t(t-1)], which decomposes into 1/(t-1) - 1/t.
✦ log|e^x - 1| - x + C
Choose the correct answer: integral of [x dx]/[(x-1)(x-2)] equals (A) log|(x-1)^2/(x-2)|+C (B) log|(x-2)^2/(x-1)|+C (C) log|((x-1)/(x-2))^2|+C (D) log|(x-1)(x-2)|+C
Hint. Decompose x/[(x-1)(x-2)] into A/(x-1)+B/(x-2) first.
Writing x=A(x-2)+B(x-1) and substituting the roots gives A=-1, B=2, so the fraction splits into -1/(x-1)+2/(x-2), integrating to -log|x-1|+2log|x-2| = log|(x-2)^2/(x-1)|.
✦ (B) log|(x-2)^2/(x-1)| + C
Choose the correct answer: integral of dx/[x(x^2+1)] equals (A) log|x|-(1/2)log(x^2+1)+C (B) log|x|+(1/2)log(x^2+1)+C (C) -log|x|+(1/2)log(x^2+1)+C (D) (1/2)log|x|+log(x^2+1)+C
Hint. Decompose 1/[x(x^2+1)] into A/x+(Bx+C)/(x^2+1) first.
Writing 1=A(x^2+1)+(Bx+C)x and matching coefficients gives A=1, B=-1, C=0, so the fraction splits into 1/x - x/(x^2+1), integrating to log|x|-(1/2)log(x^2+1).
✦ (A) log|x| - (1/2) log(x^2+1) + 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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