Question

integrate
(1+ log x)^2/x​

Answer 1

Answer:

$\int\limits^ {} \,\frac{(1+logx)^{2} }{x} dx$

Let logx=a THEN

By differentiating wrt x on both sides WE GET

(1/x) dx = da

$\int\limits \,(1+a)^{2} da\\=\int\limits \,(a^{2}+2a +1 ) da \\=\frac{a^{3} }{3}+2(\frac{a^{2}}{2} ) + a + C$

Now Substitute the value of a that we tookand that is the answer

$=\frac{logx^{3}}{3}+(logx)^{2} + logx +C$

Answer 2

Answer:

Answer:

your question is ->

you can easily solve this question with help of substitution method

put x = tanθ

differentiating both sides,

dx = sec²θ dθ

now limits : upper limits = 1 = tanθ ⇒θ = π/4 and lower limits = 0 = tanθ ⇒θ = 0°

now,

I =

I = .....(1)

we know,

so,

we know, tan(π/4- θ) = (1 - tanθ)/(1 + tanθ)

so, I =

=

= ......(2)

adding equations (1) and (2),

we get, 2I =

Step-by-step explanation: