Question

Evaluate the Integral

\(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \)

Short answer

Hence, the indefinite integration of \(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \)is sinh^-1(x/4) +c

\(\log |x + \sqrt {{x^2} + 16} | + c\)

Step-by-step solution

Step-1:- Given Data

Convert the given into\(\log |x + \sqrt {{x^2} + 16} | + c\)\(\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}} .dx\)

=\(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \)=\(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \).dx

=\(\int {\frac{1}{{\sqrt {{x^2} + {{(4)}^2}} }}} .dx\)

Here a=4

Step2:- Apply the corresponding formula

Since,\(\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}} \)=\({\sinh ^{ - 1}}({\textstyle{x \over a}})\)+c

Or\(\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}} = \log |x + \sqrt {{x^2} + {a^2}} | + c\)

\(\int {\frac{1}{{\sqrt {{x^2} + 16} }}} = \int {\frac{1}{{\sqrt {{x^2} + {{(4)}^2}} }}} .dx\)

= sinh^-1(x/4) +c

Therefore ,\(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} = {\sinh ^{ - 1}}(\frac{x}{4}) + c\)

(or)

\(\int {\frac{1}{{\sqrt {{x^2} + 16} }}} = \log |x + \sqrt {{x^2} + 16} | + c\)

Hence, the indefinite integration of \(\int {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \)is sinh^-1(x/4) +c

\(\log |x + \sqrt {{x^2} + 16} | + c\)