Question

integrate [x^2+e^logx+(e/2)^x]​

Answer 1

Answer:

I = ∫x2+elogx+(e/2)x dx

Let I1 = ∫x2 dx = x3/3 + C

I2 = ∫elogx dx

put, logx = t ,after solving we find_

I2= e2logx +C1

I3 =∫(e/2)x dx

put , (e/2)x = z

or, log(e/2)x = logz

x.log(e/2) = logz

differentiate both side,

log(e/2)dx = (1/z)dz

now, I3 = ∫{z/log(e/2).z} dz

I3 = 1/log(e/2)∫dz

I3 = z/log(e/2) + C2

I3 = (1/log(e/2)(e/2)x +C2

I = I1 + I2 + I3

I = x3/3 + e2logx+(e/2)x/log(e/2) +K

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