Question

`intsqrt(x^2 + 5)/xdx`?

Would prefer use of trig substitution if that helps. Thanks in advance

Answer 1

`int sqrt(x^2+5)/x dx = sqrt(x^2+5) +sqrt5/2 ln abs((sqrt(x^2+5)-sqrt5)/ (sqrt(x^2+5)+sqrt5))+ C`

Explanation:

Substitute:

`u = sqrt(x^2+5)`

`du = (xdx)/sqrt(x^2+5)`

so:

`int sqrt(x^2+5)/x dx = int (x^2+5)/x^2 x/sqrt(x^2+5)dx = int (u^2du)/(u^2-5)`

Reduce the degree of the numerator by splitting:

`int (u^2du)/(u^2-5) = int (u^2-5+5du)/(u^2-5) = int (1+5/(u^2-5))du`

using the linearity of the integral:

`int (u^2du)/(u^2-5) =int du +5int 1/(u^2-5)du`

`int (u^2du)/(u^2-5) =u +5int 1/(u^2-5)du`

Solve the resulting integral by partial fraction decomposition:

`1/(u^2-5) = 1/((u-sqrt5)(u+sqrt5)`

`1/(u^2-5) = A/(u-sqrt5)+B/(u+sqrt5)`

`1= A(u+sqrt5)+B(u-sqrt5)`

`1= (A+B) u + (A-B)sqrt5`

`{(A+B=0),(A-B = 1/sqrt5):}`

`{(A=1/(2sqrt5)),(B = -1/(2sqrt5)):}`

so:

`int 1/(u^2-5)du = int (du)/(2sqrt5(u-sqrt5))-int(du)/(2sqrt5(u+sqrt5))`

`int 1/(u^2-5)du = 1/(2sqrt5) ln abs(u-sqrt5)-1/(2sqrt5)ln abs (u+sqrt5)+ C`

and using the properties of logarithms:

`int 1/(u^2-5)du = 1/(2sqrt5) ln abs((u-sqrt5)/ (u+sqrt5))+ C`

Put together the partial results:

`int sqrt(x^2+5)/x dx = u +sqrt5/2 ln abs((u-sqrt5)/ (u+sqrt5))+ C`

and undo the substitution:

`int sqrt(x^2+5)/x dx = sqrt(x^2+5) +sqrt5/2 ln abs((sqrt(x^2+5)-sqrt5)/ (sqrt(x^2+5)+sqrt5))+ C`

Answer 2

Using trig substitution,as requested.
`I=sqrt(x^2+5)+sqrt5/2ln|(sqrt(x^2+5)-sqrt5)/(sqrt(x^2+5)-sqrt5)|+c`

Explanation:

`I=int(sqrt(x^2+5))/xdx`
Let, `x=sqrt5tanu=>dx=sqrt5sec^2udu`
`And,sqrt(x^2+5)=sqrt(5tan^2u+5)=sqrt5sqrt(tan^2u+1)`
`sqrt(x^2+5)=sqrt5sqrt(sec^2u)=sqrt5secu` `=>secu=sqrt(x^2+5)/sqrt5orcolor(red)(cosu=sqrt5/sqrt(x^2+5),....to(1)`
`I=int(cancelsqrt5secu)/(cancelsqrt5tanu)(sqrt5sec^2u)du`
`=sqrt5int(secu(tan^2u+1))/(tanu)du` `=sqrt5int((secutan^2u)/(tanu)+(secu)/(tanu))du`
`=sqrt5int(secutanu+(1/cosu)/(sinu/cosu))du`
`=sqrt5int(secutanu+cscu)du` `=sqrt5secu+sqrt5ln|tan(u/2)|+c` `=sqrt5secu+sqrt5/color(red)2ln|tan^color(red)2(u/2)|+c`
`=sqrt(x^2+5)+sqrt5/2ln|(1-cosu)/(1+cosu)|+c`, From (1) we get
`=sqrt(x^2+5)+sqrt5/2ln|(1-sqrt5/sqrt(x^2+5))/(1-sqrt5/sqrt(x^2+5))|+c`
`=sqrt(x^2+5)+sqrt5/2ln|(sqrt(x^2+5)-sqrt5)/(sqrt(x^2+5)-sqrt5)|+c`