Recall that indefinite integral follows the formula: `int f(x) dx = F(x) +C`
where: `f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
For the given problem `int 1/(xsqrt(9x^2+1)) dx` , it resembles one of the formula from integration table. We may apply the integral formula for rational function with roots as:
`int dx/(xsqrt(x^2+a^2))= -1/aln((a+sqrt(x^2+a^2))/x)+C`...
Recall that indefinite integral follows the formula: `int f(x) dx = F(x) +C`
where: `f(x)` as the integrand
`F(x)` as the anti-derivative function
`C` as the arbitrary constant known as constant of integration
For the given problem `int 1/(xsqrt(9x^2+1)) dx` , it resembles one of the formula from integration table. We may apply the integral formula for rational function with roots as:
`int dx/(xsqrt(x^2+a^2))= -1/aln((a+sqrt(x^2+a^2))/x)+C` .
For easier comparison, we apply u-substitution by letting: `u^2 =9x^2` or `(3x)^2` then `u = 3x ` or `u/3 =x` .
Note: The corresponding value of `a^2=1` or `1^2` then `a=1` .
For the derivative of `u` , we get: `du = 3 dx` or `(du)/3= dx` .
Plug-in the values on the integral problem, we get:
`int 1/(xsqrt(9x^2+1)) dx =int 1/((u/3)sqrt(u^2+1)) *(du)/3`
`=int 3/(usqrt(u^2+1)) *(du)/3`
`=int (du)/(usqrt(u^2+1))`
Applying the aforementioned integral formula where `a^2=1` and `a=1` , we get:
`int (du)/(usqrt(u^2+1)) =-1/1ln((1+sqrt(u^2+1))/u)+C`
` =-ln((1+sqrt(u^2+1))/u)+C`
`=ln(((1+sqrt(u^2+1))/u)^-1) + C`
`=ln(u/(1+sqrt(u^2+1))) + C`
Plug-in `u^2 =9x^2` and `u =3x` and we get the indefinite integral as:
`int 1/(xsqrt(9x^2+1)) dx=ln((3x)/(1+sqrt(9x^2+1)))+C`
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