Integrate: x sin x.
Hint. Use integration by parts with u=x (algebraic, simpler derivative) and dv=sinx dx.
With u=x, dv=sinx dx, so du=dx, v=-cosx, integration by parts gives -x cosx - integral of (-cosx)dx = -x cosx + sinx.
✦ -x cos x + sin x + C
Integrate: x sin 3x.
Hint. Use integration by parts with u=x and dv=sin3x dx.
With u=x, dv=sin3xdx, so du=dx, v=-cos3x/3, integration by parts gives -x cos3x/3 + (1/3) integral of cos3x dx.
✦ -x cos 3x/3 + (1/9) sin 3x + C
Integrate: x^2 e^x.
Hint. Use integration by parts with u=x^2 (reduces the power each time) and dv=e^x dx, applying the technique twice.
With u=x^2, dv=e^xdx, the first pass gives x^2e^x - 2 integral of x e^x dx. Applying integration by parts again to x e^x dx (now u=x) gives xe^x-e^x, combining to the final result.
✦ e^x (x^2 - 2x + 2) + C
Integrate: x log x.
Hint. Use integration by parts with u=log x (its derivative 1/x is simpler) and dv=x dx.
With u=logx, dv=xdx, so du=dx/x, v=x^2/2, integration by parts gives (x^2/2)logx - integral of (x/2)dx.
✦ (x^2/2) log x - x^2/4 + C
Integrate: x log 2x.
Hint. Use integration by parts with u=log 2x and dv=x dx, same structure as x log x.
With u=log2x, dv=xdx, so du=dx/x (since the derivative of log2x is still 1/x), v=x^2/2, integration by parts gives (x^2/2)log2x - integral of (x/2)dx.
✦ (x^2/2) log 2x - x^2/4 + C
Integrate: x^2 log x.
Hint. Use integration by parts with u=log x and dv=x^2 dx.
With u=logx, dv=x^2dx, so du=dx/x, v=x^3/3, integration by parts gives (x^3/3)logx - integral of (x^2/3)dx.
✦ (x^3/3) log x - x^3/9 + C
Integrate: x sin^-1 x.
Hint. Use integration by parts with u=sin^-1 x and dv=x dx, then simplify the resulting integral of x^2/sqrt(1-x^2).
With u=sin^-1x, dv=xdx, v=x^2/2, integration by parts leaves integral of x^2/sqrt(1-x^2) dx; since x^2=-(1-x^2)+1, this splits into two standard forms that integrate directly.
✦ (1/4)(2x^2-1) sin^-1 x + (x/4) sqrt(1-x^2) + C
Integrate: x tan^-1 x.
Hint. Use integration by parts with u=tan^-1 x and dv=x dx, then simplify x^2/(1+x^2) by writing it as 1-1/(1+x^2).
With u=tan^-1x, dv=xdx, v=x^2/2, integration by parts leaves integral of x^2/(1+x^2)dx = integral of [1-1/(1+x^2)]dx = x-tan^-1x.
✦ (1/2)(x^2+1) tan^-1 x - x/2 + C
Integrate: x cos^-1 x.
Hint. Use integration by parts with u=cos^-1 x and dv=x dx, then handle the resulting x^2/sqrt(1-x^2) integral as in the sin^-1 case.
With u=cos^-1x, dv=xdx, v=x^2/2, integration by parts leaves +integral of x^2/sqrt(1-x^2)dx (the sign flips since d/dx(cos^-1x)=-1/sqrt(1-x^2)), which splits into standard forms as before.
✦ (x^2/2) cos^-1 x - (x/4) sqrt(1-x^2) + (1/4) sin^-1 x + C
Integrate: (sin^-1 x)^2.
Hint. Use integration by parts with u=(sin^-1 x)^2 and dv=dx, then evaluate the resulting integral via the substitution w=sin^-1 x.
With u=(sin^-1x)^2, dv=dx, v=x, integration by parts leaves -2 integral of x sin^-1x/sqrt(1-x^2)dx. Substituting w=sin^-1x (so x=sinw) turns this into integral of w sinw dw, itself another by-parts integral.
✦ x (sin^-1 x)^2 - 2x + 2 sqrt(1-x^2) sin^-1 x + C
Integrate: (x cos^-1 x)/sqrt(1-x^2).
Hint. Notice x/sqrt(1-x^2) dx is exactly -d[sqrt(1-x^2)], so integrate by parts with v=sqrt(1-x^2) directly.
Recognising x/sqrt(1-x^2)dx = -d(sqrt(1-x^2)), integration by parts with u=cos^-1x and v=sqrt(1-x^2) gives -cos^-1x sqrt(1-x^2) - integral of dx (the sign works out since d/dx(cos^-1x)=-1/sqrt(1-x^2)).
✦ -x - sqrt(1-x^2) cos^-1 x + C
Integrate: x sec^2 x.
Hint. Use integration by parts with u=x and dv=sec^2x dx, since v=tanx is a known antiderivative.
With u=x, dv=sec^2xdx, so v=tanx, integration by parts gives x tanx - integral of tanx dx, and integral of tanx dx = -log|cosx| = log|secx|.
✦ x tan x + log|cos x| + C
Integrate: tan^-1 x.
Hint. Use integration by parts with u=tan^-1 x and dv=dx (treating this as 1 times tan^-1x).
With u=tan^-1x, dv=dx, v=x, integration by parts gives x tan^-1x - integral of x/(1+x^2)dx, and the remaining integral is a standard log form via t=1+x^2.
✦ x tan^-1 x - (1/2) log(1+x^2) + C
Integrate: x(log x)^2.
Hint. Use integration by parts with u=(log x)^2 and dv=x dx, reducing to the already-solved integral of x log x.
With u=(logx)^2, dv=xdx, v=x^2/2, integration by parts leaves minus the integral of x logx dx; since that is the standard result (x^2/2)logx-x^2/4 from a previous question, substituting it back completes the answer.
✦ (x^2/2)(log x)^2 - (x^2/2) log x + x^2/4 + C
Integrate: (x^2+1) log x.
Hint. Split into integral of x^2 logx plus integral of logx, and apply integration by parts to each separately.
The first piece x^2logx integrates by parts to (x^3/3)logx-x^3/9, and the second piece logx integrates by parts (with dv=dx) to x logx - x; adding gives the combined result.
✦ (x^3/3) log x - x^3/9 + x log x - x + C
Integrate: e^x(sin x + cos x).
Hint. Recognise the special form integral of e^x[f(x)+f'(x)]dx = e^x f(x)+C with f(x)=sin x.
Since f(x)=sinx has f'(x)=cosx, the integrand matches e^x[f(x)+f'(x)] exactly, so the antiderivative is simply e^x f(x)=e^x sinx.
✦ e^x sin x + C
Integrate: [x e^x]/(1+x)^2.
Hint. Rewrite x/(1+x)^2 as 1/(1+x) - 1/(1+x)^2, and recognise this as f(x)+f'(x) for f(x)=1/(1+x).
Since f(x)=1/(1+x) has f'(x)=-1/(1+x)^2, and x/(1+x)^2=1/(1+x)-1/(1+x)^2 matches f(x)+f'(x) exactly, the antiderivative is e^x f(x)=e^x/(1+x).
✦ e^x/(1+x) + C
Integrate: e^x(1+sin x)/(1+cos x).
Hint. Use the half-angle identities 1+cosx=2cos^2(x/2) and sinx=2sin(x/2)cos(x/2) to rewrite the fraction as f(x)+f'(x) for f(x)=tan(x/2).
Rewriting, (1+sinx)/(1+cosx) = (1/2)sec^2(x/2)+tan(x/2), which matches f(x)+f'(x) exactly for f(x)=tan(x/2), so the antiderivative is e^x tan(x/2).
✦ e^x tan(x/2) + C
Integrate: e^x(1/x - 1/x^2).
Hint. Recognise this as f(x)+f'(x) for f(x)=1/x, since f'(x)=-1/x^2.
Since f(x)=1/x has f'(x)=-1/x^2, the integrand matches e^x[f(x)+f'(x)] exactly, so the antiderivative is e^x f(x)=e^x/x.
✦ e^x/x + C
Integrate: [(x-3)e^x]/(x-1)^3.
Hint. Write x-3=(x-1)-2, splitting the fraction into 1/(x-1)^2 - 2/(x-1)^3, and recognise this as f(x)+f'(x) for f(x)=1/(x-1)^2.
Since f(x)=1/(x-1)^2 has f'(x)=-2/(x-1)^3, and (x-3)/(x-1)^3=1/(x-1)^2-2/(x-1)^3 matches f(x)+f'(x) exactly, the antiderivative is e^x f(x)=e^x/(x-1)^2.
✦ e^x/(x-1)^2 + C
Integrate: e^(2x) sin x.
Hint. Apply integration by parts twice, which produces a multiple of the original integral, then solve for it algebraically.
Applying integration by parts twice (first with u=sinx, dv=e^(2x)dx, then again on the resulting cosine integral) gives an equation I=-e^(2x)cosx+2e^(2x)sinx-4I, which solves to 5I=e^(2x)(2sinx-cosx).
✦ e^(2x) (2sin x - cos x)/5 + C
Integrate: sin^-1(2x/(1+x^2)).
Hint. Recognise 2x/(1+x^2) as sin(2 tan^-1 x), a standard identity, which simplifies the inverse sine dramatically.
Since 2x/(1+x^2) = sin(2tan^-1x) for x in the relevant range, sin^-1(2x/(1+x^2)) = 2tan^-1x, reducing the problem to the already-solved integral of tan^-1x (doubled).
✦ 2x tan^-1 x - log(1+x^2) + C
Choose the correct answer: integral of x^2 e^(x^3) dx equals (A) (1/3)e^(x^3)+C (B) (1/3)e^(x^2)+C (C) (1/2)e^(x^3)+C (D) (1/2)e^(x^2)+C
Hint. Put t=x^3, so dt=3x^2 dx, exactly the leftover factor.
With t=x^3, dt=3x^2dx, the integral becomes (1/3) integral of e^t dt = (1/3)e^t.
✦ (A) (1/3) e^(x^3) + C
Choose the correct answer: integral of e^x sec x (1+tan x) dx equals (A) e^x cos x+C (B) e^x sec x+C (C) e^x sin x+C (D) e^x tan x+C
Hint. Expand the product and recognise it as f(x)+f'(x) for f(x)=secx.
Expanding, e^x secx(1+tanx) = e^x(secx+secxtanx), and since f(x)=secx has f'(x)=secxtanx, this matches e^x[f(x)+f'(x)] exactly, giving e^x f(x)=e^x secx.
✦ (B) e^x sec x + 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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