We will calculate ∫((x2+16))dx/(x4)
put x=4tan(t)
So dx=4sec2tdt
Now integration becomes ∫((4(1+tan2t))∗4sec2t)dt/(44tan4(t))
=∫(sec3(t)dt)/(16tan4(t))
=∫(1/cos3(t))∗(16(cos4(t)/sin4(t)))
=∫(cos(t))/(16sin4(t))
Now let sin(t)=z So cos(t)dt=dz
So integration becomes ∫(dz/16z4)=∫((z(−4)dz))/16
= -1/(48z^3)
=-1/(48sin^3(t)) As ∫(xndx)=(x(n+1))/(n+1) We had taken x=4tan(t) ,so tan(t)=x/4 ,in a right angle triangle tan(t)=p/b , so p=x & b=4
So h=(x2+16) ,So sin(t)=p/h=x/((x2+16)) =−(((x2+16))3)/(48x3)+C (ans)
Here C is constant of integration.
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