`int arctan(4t) dt` Evaluate the integral

`intarctan(4t)dt`

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Using the above integration by parts method,

`intarctan(4t)dt=arctan(4t)*int1dt-int(d/dt(arctan(4t)int1dt)dt`

`=arctan(4t)*t-int(4/((4t)^2+1)*t)dt`

`=tarctan(4t)-4intt/(16t^2+1)dt`

Now let's evaluate `intt/(16t^2+1)dt`  by using the method of substitution,

Substitute `x=16t^2+1,=>dx=32tdt`

`intt/(16t^2+1)dt=intdx/(32x)`

`=1/32ln|x|`

substitute back `x=16t^2+1`

`=1/32ln|16t^2+1|`

`intarctan(4t)=t*arctan(4t)-4/32ln|16t^2+1|+C`

`intarctan(4t)=t*arctan(4t)-1/8ln|16t^2+1|+C`

C is a constant.

`intarctan(4t)dt`

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Using the above integration by parts method,

`intarctan(4t)dt=arctan(4t)*int1dt-int(d/dt(arctan(4t)int1dt)dt`

`=arctan(4t)*t-int(4/((4t)^2+1)*t)dt`

`=tarctan(4t)-4intt/(16t^2+1)dt`

Now let's evaluate `intt/(16t^2+1)dt`  by using the method of substitution,

Substitute `x=16t^2+1,=>dx=32tdt`

`intt/(16t^2+1)dt=intdx/(32x)`

`=1/32ln|x|`

substitute back `x=16t^2+1`

`=1/32ln|16t^2+1|`

`intarctan(4t)=t*arctan(4t)-4/32ln|16t^2+1|+C`

`intarctan(4t)=t*arctan(4t)-1/8ln|16t^2+1|+C`

C is a constant.