Question

integrate dx/4xsqrt(x^2-16)

Answer 1

take
\[x^2-16=z^2 \\so \\differentiate ~~\it ~~both ~~sides\\2xdx=2zdz \rightarrow xdx=zdz\]
so u have
\[I=\int\limits_{}^{} \frac{ dx }{ 4x \sqrt{x^2-16} } \\now~multiply ~~x ~~with ~~numerator~~and~~denominator\\I=\int\limits_{}^{}\frac{ x dx }{ 4x^2 \sqrt{16-x^2} } \]
now just substitute
xdx=zdz and
\[x^2=16+z^2\]
and u'll have
\[I=\int\limits_{}^{}\frac{ z dz }{ 4(z^2+16)*z }=\frac{ 1 }{ 4 } \int\limits_{}^{}\frac{ dz }{z^2+16 }=\frac{ 1 }{ 4 } *[\frac{ 1 }{4 } \tan^{-1} \frac{ z }{ 4 }]+C\]
now just substitute z in terms of x and that'll be all