Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 11

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.