Integrate: 2x/(1+x^2).
Hint. Put t=1+x^2, so dt=2x dx — the numerator is exactly dt.
With t=1+x^2, dt=2x dx, so the integral becomes integral of dt/t = log|t|. Since 1+x^2 is always positive, the absolute value can be dropped.
✦ log(1+x^2) + C
Integrate: (log x)^2/x.
Hint. Put t=log x, so dt=dx/x.
With t=logx, dt=dx/x, the integral becomes integral of t^2 dt = t^3/3, giving the answer once t is replaced back by logx.
✦ (log x)^3 / 3 + C
Integrate: 1/(x+x log x).
Hint. Factor x out of the denominator first, then substitute for the bracket.
Factoring, x+x logx = x(1+logx), so with t=1+logx, dt=dx/x, the integral becomes integral of dt/t = log|t|.
✦ log|1+log x| + C
Integrate: sin x . sin(cos x).
Hint. Put t=cos x, so dt=-sin x dx.
With t=cosx, dt=-sinx dx, the integral becomes -integral of sin t dt = cos t, giving cos(cosx) once substituted back.
✦ cos(cos x) + C
Integrate: sin(ax+b) cos(ax+b).
Hint. Use sin(2theta)=2 sin(theta)cos(theta) to rewrite the product as a single sine.
Since sin(ax+b)cos(ax+b) = (1/2)sin(2ax+2b), integrating this scaled sine gives the result, with the extra factor of 2a coming from the chain rule on the inner argument.
✦ -cos(2ax+2b) / (4a) + C
Integrate: sqrt(ax+b).
Hint. Put t=ax+b, so dt=a dx, then apply the power rule.
With t=ax+b, dt=a dx, the integral becomes (1/a) integral of t^(1/2) dt = (1/a)(2/3)t^(3/2), giving the result once t is replaced back.
✦ (2/(3a)) (ax+b)^(3/2) + C
Integrate: x sqrt(x+2).
Hint. Put t=x+2 so x=t-2, then expand the product before integrating each power of t.
With t=x+2, x=t-2, so x sqrt(x+2) dx = (t-2)sqrt(t) dt = (t^(3/2)-2t^(1/2))dt, and integrating each term gives the result in terms of t, then back in x.
✦ (2/5)(x+2)^(5/2) - (4/3)(x+2)^(3/2) + C
Integrate: x sqrt(1+2x^2).
Hint. Put t=1+2x^2, so dt=4x dx.
With t=1+2x^2, dt=4x dx, the integral becomes (1/4) integral of sqrt(t) dt = (1/4)(2/3)t^(3/2), giving the answer once t is replaced back.
✦ (1/6)(1+2x^2)^(3/2) + C
Integrate: (4x+2) sqrt(x^2+x+1).
Hint. Put t=x^2+x+1, so dt=(2x+1)dx, and note (4x+2)=2(2x+1) is exactly a multiple of dt.
With t=x^2+x+1, dt=(2x+1)dx, the numerator (4x+2)dx equals 2dt, so the integral becomes 2 integral of sqrt(t) dt = 2(2/3)t^(3/2).
✦ (4/3)(x^2+x+1)^(3/2) + C
Integrate: 1/(x-sqrt(x)).
Hint. Factor sqrt(x) out of the denominator, then substitute t=sqrt(x).
Since x-sqrt(x) = sqrt(x)(sqrt(x)-1), put t=sqrt(x) so dx=2t dt; the integral becomes integral of 2t/[t(t-1)]dt = 2 integral of dt/(t-1) = 2log|t-1|.
✦ 2 log|sqrt(x) - 1| + C
Integrate: x/sqrt(x+4).
Hint. Put t=sqrt(x+4), so x=t^2-4 and dx=2t dt, turning the square root into a plain polynomial in t.
With t=sqrt(x+4), x=t^2-4, dx=2t dt, the integral becomes integral of (t^2-4)/t . 2t dt = 2 integral of (t^2-4) dt = 2(t^3/3-4t), which simplifies back in terms of x.
✦ (2/3)(x-8) sqrt(x+4) + C
Integrate: (x^3-1)^(1/3) x^5.
Hint. Put t=x^3-1 so x^3=t+1 and dt=3x^2 dx; write x^5=x^3.x^2=(t+1)x^2 to isolate the dt piece.
With t=x^3-1, dt=3x^2dx, write x^5 dx = (t+1)x^2 dx = (t+1)dt/3, so the integral becomes (1/3) integral of t^(1/3)(t+1) dt = (1/3)[(3/7)t^(7/3)+(3/4)t^(4/3)].
✦ (1/7)(x^3-1)^(7/3) + (1/4)(x^3-1)^(4/3) + C
Integrate: x^2/(2+3x^3)^3.
Hint. Put t=2+3x^3, so dt=9x^2 dx.
With t=2+3x^3, dt=9x^2dx, the integral becomes (1/9) integral of t^-3 dt = (1/9)(-1/(2t^2)), giving the answer once t is replaced back.
✦ -1/(18(2+3x^3)^2) + C
Integrate: 1/(x(log x)^m), x>0, m not 1.
Hint. Put t=log x, so dt=dx/x, reducing this to a pure power of t.
Putting t=logx gives dt=dx/x, so the integral becomes integral of t^-m dt = t^(1-m)/(1-m), valid since m is not 1, giving the answer once t is replaced back by logx.
✦ (log x)^(1-m) / (1-m) + C
Integrate: x/(9-4x^2).
Hint. Put t=9-4x^2, so dt=-8x dx.
With t=9-4x^2, dt=-8x dx, the integral becomes -(1/8) integral of dt/t = -(1/8)log|t|.
✦ -(1/8) log|9-4x^2| + C
Integrate: e^(2x+3).
Hint. Put t=2x+3, so dt=2dx.
With t=2x+3, dt=2dx, the integral becomes (1/2) integral of e^t dt = (1/2)e^t, giving the answer once t is replaced back.
✦ (1/2) e^(2x+3) + C
Integrate: x/e^(x^2).
Hint. Write the integrand as x e^(-x^2) and put t=x^2, so dt=2x dx.
Rewriting as x e^-x^2, put t=x^2, dt=2xdx, so the integral becomes -(1/2) integral of e^-t dt = -(1/2)e^-t, giving the result once t is replaced back.
✦ -1/(2 e^(x^2)) + C
Integrate: e^(tan^-1 x)/(1+x^2).
Hint. Put t=tan^-1 x, so dt=dx/(1+x^2) — the denominator is exactly dt.
With t=tan^-1x, dt=dx/(1+x^2), the integral becomes integral of e^t dt = e^t, giving the answer once t is replaced back.
✦ e^(tan^-1 x) + C
Integrate: (e^(2x)-1)/(e^(2x)+1).
Hint. Divide numerator and denominator by e^x first, then substitute for the resulting sum.
Dividing through by e^x gives (e^x-e^-x)/(e^x+e^-x). Putting t=e^x+e^-x, dt=(e^x-e^-x)dx, the integral becomes integral of dt/t = log|t|, which is positive so the bars can be dropped.
✦ log(e^x + e^(-x)) + C
Integrate: (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)).
Hint. Put t equal to the denominator, since its derivative is proportional to the numerator.
With t=e^(2x)+e^(-2x), dt=2(e^(2x)-e^(-2x))dx, the integral becomes (1/2) integral of dt/t = (1/2)log|t|, which is positive so bars can be dropped.
✦ (1/2) log(e^(2x) + e^(-2x)) + C
Integrate: tan^2(2x-3).
Hint. Rewrite using tan^2(theta)=sec^2(theta)-1, then integrate term by term.
Since tan^2(2x-3)=sec^2(2x-3)-1, the integral splits into integral of sec^2(2x-3), which gives (1/2)tan(2x-3) by the chain rule, minus integral of 1, which gives x.
✦ (1/2) tan(2x-3) - x + C
Integrate: sec^2(7-4x).
Hint. Put t=7-4x, so dt=-4dx.
With t=7-4x, dt=-4dx, the integral becomes -(1/4) integral of sec^2t dt = -(1/4)tan t, giving the answer once t is replaced back.
✦ -(1/4) tan(7-4x) + C
Integrate: sin^-1 x / sqrt(1-x^2).
Hint. Put t=sin^-1 x, so dt=dx/sqrt(1-x^2) — the denominator is exactly dt.
With t=sin^-1x, dt=dx/sqrt(1-x^2), the integral becomes integral of t dt = t^2/2, giving the answer once t is replaced back.
✦ (sin^-1 x)^2 / 2 + C
Integrate: (2cos x - 3sin x)/(6cos x + 4sin x).
Hint. Write the numerator as a combination of the denominator and its derivative, then split the integral.
The derivative of (6cosx+4sinx) is (4cosx-6sinx). Solving 2cosx-3sinx = A(6cosx+4sinx)+B(4cosx-6sinx) gives A=0, B=1/2, so the integral reduces to (1/2) integral of (denominator)'/(denominator) dx = (1/2)log|denominator|.
✦ (1/2) log|6cos x + 4sin x| + C
Integrate: 1/(cos^2 x (1-tan x)^2).
Hint. Put t=tan x, so dt=sec^2 x dx = dx/cos^2 x — exactly the leftover factor.
With t=tanx, dt=dx/cos^2x, the integral becomes integral of dt/(1-t)^2 = 1/(1-t), giving the answer once t is replaced back by tanx.
✦ 1/(1 - tan x) + C
Integrate: cos(sqrt(x))/sqrt(x).
Hint. Put t=sqrt(x), so dt=dx/(2 sqrt(x)).
With t=sqrt(x), dt=dx/(2sqrt(x)), the integral becomes 2 integral of cos t dt = 2 sin t, giving the answer once t is replaced back.
✦ 2 sin(sqrt(x)) + C
Integrate: sqrt(sin 2x) cos 2x.
Hint. Put t=sin 2x, so dt=2cos 2x dx.
With t=sin2x, dt=2cos2x dx, the integral becomes (1/2) integral of sqrt(t) dt = (1/2)(2/3)t^(3/2), giving the answer once t is replaced back.
✦ (1/3)(sin 2x)^(3/2) + C
Integrate: cos x / sqrt(1+sin x).
Hint. Put t=1+sin x, so dt=cos x dx.
With t=1+sinx, dt=cosx dx, the integral becomes integral of t^(-1/2) dt = 2t^(1/2), giving the answer once t is replaced back.
✦ 2 sqrt(1+sin x) + C
Integrate: cot x . log(sin x).
Hint. Put t=log(sin x), so dt=cot x dx — since d/dx[log(sinx)]=cosx/sinx=cotx.
With t=log(sinx), dt=cotx dx, the integral becomes integral of t dt = t^2/2, giving the answer once t is replaced back.
✦ (log(sin x))^2 / 2 + C
Integrate: sin x / (1+cos x).
Hint. Put t=1+cos x, so dt=-sin x dx.
With t=1+cosx, dt=-sinx dx, the integral becomes -integral of dt/t = -log|t|, giving the answer once t is replaced back.
✦ -log|1+cos x| + C
Integrate: sin x / (1+cos x)^2.
Hint. Put t=1+cos x, so dt=-sin x dx.
Putting t=1+cosx gives dt=-sinx dx, so the integral becomes -integral of t^-2 dt = 1/t, giving the answer once t is replaced back.
✦ 1/(1+cos x) + C
Integrate: 1/(1+cot x).
Hint. Rewrite as sin x/(sin x+cos x), then split the numerator into the denominator plus its derivative.
Since the derivative of sinx+cosx is cosx-sinx, sinx can be written as (1/2)(sinx+cosx)+(1/2)(cosx-sinx), so the integral splits into (1/2) integral of dx minus (1/2) integral of (denominator)'/(denominator) dx.
✦ x/2 - (1/2) log|sin x + cos x| + C
Integrate: 1/(1-tan x).
Hint. Rewrite as cos x/(cos x - sin x), then split the numerator into the denominator plus its derivative.
Writing cosx = (1/2)(cosx-sinx) + (1/2)(sinx+cosx) [matching the denominator and its derivative, since the derivative of cosx-sinx is -sinx-cosx], the integral splits into (1/2) integral of dx minus (1/2) integral of (denominator)'/(denominator) dx.
✦ x/2 - (1/2) log|cos x - sin x| + C
Integrate: sqrt(tan x) / (sin x cos x).
Hint. Multiply numerator and denominator by sec^2 x so the denominator becomes tan x, then substitute t=tan x.
Multiplying through by sec^2x/sec^2x, the denominator sinxcosx.sec^2x simplifies to tanx, so with t=tanx, dt=sec^2xdx, the integral becomes integral of t^(-1/2)dt = 2 t^(1/2).
✦ 2 sqrt(tan x) + C
Integrate: (1+log x)^2 / x.
Hint. Put t=1+log x, so dt=dx/x.
With t=1+logx, dt=dx/x, the integral becomes integral of t^2 dt = t^3/3, giving the answer once t is replaced back.
✦ (1+log x)^3 / 3 + C
Integrate: (x+1)(x+log x)^2 / x.
Hint. Rewrite (x+1)/x as (1+1/x), which is exactly the derivative of the bracket (x+log x).
Since (x+1)/x = 1+1/x = d/dx(x+logx), put t=x+logx, dt=(1+1/x)dx, and the integral becomes integral of t^2 dt = t^3/3.
✦ (x+log x)^3 / 3 + C
Integrate: x^3 sin(tan^-1(x^4)) / (1+x^8).
Hint. Put u=x^4 first (so 1+x^8=1+u^2), then v=tan^-1 u to reduce this to a simple sine integral.
With u=x^4, du=4x^3dx, the integral becomes (1/4) integral of sin(tan^-1u)/(1+u^2) du. Now put v=tan^-1u, dv=du/(1+u^2), giving (1/4) integral of sin v dv = -(1/4)cos v.
✦ -(1/4) cos(tan^-1(x^4)) + C
Choose the correct answer: integral of (10x^9+10^x loge10)/(x^10+10^x) dx equals (A) 10^x-x^10+C (B) 10^x+x^10+C (C) (10^x-x^10)^-1+C (D) log(10^x+x^10)+C
Hint. Notice the numerator is exactly the derivative of the denominator.
Since d/dx(x^10+10^x) = 10x^9+10^x log_e10, which is exactly the numerator, the integral is of the form (denominator)'/(denominator), giving log|denominator|.
✦ (D) log(10^x + x^10) + C
Choose the correct answer: integral of dx/(sin^2 x cos^2 x) equals (A) tan x+cot x+C (B) tan x-cot x+C (C) tan x cot x+C (D) tan x-cot 2x+C
Hint. Write sin^2x+cos^2x=1 in the numerator, split the fraction, and integrate each piece.
Since 1/(sin^2xcos^2x) = (sin^2x+cos^2x)/(sin^2xcos^2x) = sec^2x+cosec^2x, integrating sec^2x gives tanx and integrating cosec^2x gives -cotx, so adding these two pieces gives tanx-cotx.
✦ (B) tan x - cot 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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