In questions 4-6 show all workings as in the form of a table indicated and apply integration by parts as indicated by the formula ∫udv=uv−∫vdu or similar. I=∫tan −1
(3x−1)dx Let u=3x−1 then dw
dx
​
=3 So we have ∫tan −1
(3x−1)dx= 3
1
​
∫tan −1
udu= 3
1
​
I 2
​
For I 2
​
w=…dv=…
du
dw
​
=…v=…
∫tan −1
udu=wv−∫vdw
​
Then the answer will be =
​
∫tan −1
(3x−1)dx
(…)tan −1
(3x−1)− constant ln(…)+c
​
I=∫x 2
cos( 2
x
​
)dx Let u=…
dx
du
​
=…
​
dv=cos( 2
x
​
)dx
v=…
​
(for v use substitution w= 2
z
​
and dx
dw
​
= 2
1
​
) I
​
=∫x 2
cos 2
x
​
)dx
=(…)−4∫xsin 2
x
​
dx
=(…)−4I 2
​
​
Let I 2
​
=∫xsin 2
π
​
dx Then u=…
ds
dt
​
=…
​
dv=…
v=…
​
(using substitution w= 2
π
​
and dx
dw
​
= 2
1
​
to obtain v ). We have (answer worked out)