how do i integrate the integral [x+(7/58)]/[x^2+9] ?

Make a u-substitution for x^2 + 9

( [ x + ( 7 / 58 ) ] / [ x^2 + 9 ] ) dx

[ ( x + ( 7 / 58 ) ) / u ] * dx <-- substitute u

u = x^2 + 9

du = 2x dx < integrate both sides

dx = 1 / 2x du < solve for dx

now plug 1 / 2x in place of dx from the equation from up top.

[ x + ( 7 / 58 ) ] / u ] * 1 / 2x du < This will cause the x to drop out.

[ ( 7 / 58 ) / u ] * ( 1 / 2 ) < clean up

( 7 / 116u ) < now find the antiderivative.

7 / 116 ln u < undo the u substitution

7 / 116 ln (x^2 + 9)

First separate into elementary integrable expressions:

That is here to find a expression where the nominator is the derivative of the denominator plus a constant divided by the denominator.

∫ (x+(7/58))/(x²+9) dx

= 1/2 · ∫ (2x+7/29)/(x²+9) dx

= 1/2· ∫ (2x)/(x²+9) dx + 7/58 · ∫ 1/(x²+9) dx

For the first integral the rule

∫ f'(x)/f(x) dx = ln (f(x)) applies. Hence

1/2 · ∫ (2x)/(x²+9) dx = 1/2 ·ln(x²+9)

The second integral could be transformed in an integral of the type:

∫ 1/ (x²+1) dx = arctan(x)

Here

7/58 · ∫ 1/(x²+9) dx

= 7/522 · ∫ 1/((x/3)²+1) dx

with the substitution u = (x/3) => dx=3 du

= 7/174 · ∫ 1/ (u²+1) du

= 7/174 · arctan(u)

= 7/174 · arctan(x/3)

Therefore the result is:

∫ (x-7/58)/(x²+9) dx = 1/2 ·ln(x²+9) + 7/174 ·arctan(x/3) + C