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dy/dx=1/(x+1)(y+1) (y+1)dy=dx/x+1 y^2/2+y=ln(x+1)+C
Substitute x= sin 2 t dx=d[1/2(1-cos t)] Change the limits accordingly... Apply Integration By parts...
=-cosx+c
int[log(x+sqrt(1+x^2))]/[sqrt(1+x^2)] dx use integration by substitution twice. first substitution, let u = x + sqrt(1+x^2) then du = [1 + x/sqrt(1+x^2)]dx = [(sqrt(1+x^2) + x]/sqrt(1+x^2) dx =...
The question is ∫ tan -1 [ 1/(x 2 +x+1) ] dx This can be written as ∫ tan -1 [ {(x+ 1)-x} / {1+x(x+1)} ] dx From the formula of tan -1 [ (m-n) / 1+mn ], We get ∫ tan -1 (x+1) – tan -1 x dx I guess now...
It is a lame joke. “bo” comes out as a constant and dy is integrated as “y” therefore- integration of body is “boy”
(-2n 3 -6n 2 -4n).ln(x+n)+[12x 5 +(15n-90)x 4 +(260-60n)x 3 -360x 2 +(120n 2 +360n+240)x]/60
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integrate x!/(x+n)! ?
Hi, See the grapgh of both tan^-11 and cot^-1 Now the integrations will be equal till the time gratest integer is giving zero. But cot^-1 starts from pi/2 and decreses and tan^-1 starts from 0 and...
Hi The answer is zero. As the function is odd fucntion so its integration on both sides of y axix will give us equal values but of opposite sign to that of on other side of y axis. So by the peoperty...
Hello Student, Just take lnx = t So 1/x dx = dt So ques becomes integration sint dt...
Hi, Take Ln X = t So by this substitution dt is available which is 1/x. Now our function has reduced to Sin t . dt which is easily integrable. After integration do the back substitution to get the...
what is your question? please post your doubt/s. .................... .................................... ........................................................
Hi, Apply the bi part form contineously, ull see that the last integration will also become integrable where the first ones are integrable already. So you can reach to the answer. By applying...
You can this in 2 ways: either along x axis which would ease the situation as the area would be calculated using the even function property of the integrable area, i.e. same magnitude of area on both...
Hi, to solve this integration, put x = tan y so, dx = sec 2 y dy, and the limits of y would be - /4 to /4 Now the integral becomes, [1/sec 2 y] * sec 2 y dy = thanks
Obtain the integration of Sin 3 x from the Sin3x equality involving Sin 3 x in terms of Sinx and Sin 3x. Now, for∫x 2 sin 3 x dx, use by parts (with ILATE: Inverse, log, algebraic, Trig,...
Please check your answer then. Applying the property of even function here will be helpful. That could be a case I can expect that you have integrated the whole area in one integrand which could have...
Hi, sinx dx + cosx dx [-cosx] 90 0 + [sinx] 90 0 = 1+1 = 2...
Hi, Assume sinx = t, cosx dx = dt, so, integration becomes 1 / (1-t 2 ) dt Solove this integral by writing (1-t 2 ) = (1-t) (1+t) thnaks.
The result would be a function of [log(x+7)] 3 -2 by applying the limits, upper and lower.......... It would be easy for you know to to put the limits and solve accordingly as a logarthimic...
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