integrate the following...?

Question

integrate: arctan(7z)dz

Answer 1

♦ =x* atan(7x) –ln(1+(7x)^2) /14 +C;
switch on your cap-carrier;

Answer 2

First, let's put u = 7z, z = u/7 dz = 1/7 du.
Then the integral becomes
1/7* ∫ arctan u du.
Next, we do the last integral by parts.
Let
U = arctan u dV = du
dU = du/(1+u²) V = u
So we have
1/7∫ arctan u du = 1/7*[u arctan u -½∫ 2u du/(1+u²) ] =
1/7*[ u arctan u - ½ ln(1+u²) ].
Since u = 7z, our final result is
z arctan 7z - 1/14*ln(1 + 49z²) + C.

Answer 3

integral arctan(7z)dz
= 7z arctan7z - .5ln(1+49z^2) + C

Answer 4

use an integral table or the ti-89

[integral] arctan x dx = x arctan x - (1/2) ln(1+x^{2}) + C

so integrate: arctan(7z)dz

7z arctan 7z - (1/2) ln(1+49z^{2}) + C

Answer 5

I think it gets easier if we first find the integral of arctan(x). Put x = tan(t), so that dx = sec^2(t) dt. So, we get

Int t sec^2(t) dt Now, put u = t and dv = sec^2(t) dt, so that v = tan(t) and du = dt.

By parts, Int t sec^2(t) dt = t tan(t) - Int tan(t) dt
We know Int tan(t) dt = Int sin(t)/cos(t) dt = - ln(cos(t).

So, Int t sec^2(t) dt = t tan(t) - ln(cos(t) + C.
Since x = tan(t), t = arctan(x). sec^2(t) = 1/cos^2(t) = 1 + tan^2(t) = 1 + x^2, so that cos(t) = 1/(1 + x^2)^(1/2). It follows that

Int arctan(x) dx = x arctan(x) - ln((1+ x^2)^(-1/2) + C = x arctan(x) + (1/2) ln(1+ x^2) + C

Now, Int arctan(7z)dz = 1/7 Int arctan(7z) 7 dz= (1/7) [7z arctan(7z) + (1/2) ln(1+ 49z^2)] + C

Answer 6

You can only accomplish that by partial integration :
Integ(f'g) = [fg] - Integ(fg')
Take f'=dz => f = z
Take g=arctan(7z) => g' = 7/(1+49z²)
So we have : [z arctan(7z)] - Integ(7z dz/(1+49z²))
and the last integral is Integ((1/14)d(1+49z²)/(1+49z²))
= 1/14 log(1+49z²)
So the soluton is : z arctan(7z) - log(1+49z²)/14.

Answer 7

Integration by parts:
let u=arctan(7z), then dv = dz.
Therefore,
du = 7dz/(1+49z^2)
v = z

The initial integral =
z arctan(7z)-∫7zdz/(1+49z^2)

To find ∫7zdz/(1+49z^2),
substitute t=1+49z^2.
Then dt=98zdz and 7zdz=(1/14)dt
Finally we get
∫7zdz/(1+49z^2)
=(1/14)∫dt/t
=(1/14) ln|t| + C
=(1/14) ln(1+49z^2) + C

The final answer is:
z arctan(7z) - (1/14) ln(1+49z^2) + C