How do you integrate by substitution $\int [x^{2} + \frac{1}{{(3 x)}^{2}}] d x$ $int [x^2+1/(3x)^2]dx$ ?

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Answer 1 · 2017-01-05T16:29:43

$int \ x^2 + 1/(3x)^2 \ dx = x^3/3 - 1/(9x) +c$

You do not need to use a substitution, just the power rule, as:

$int \ x^2 + 1/(3x)^2 \ dx = int \ x^2 + 1/(9x^2) \ dx$
$" "= int \ x^2 + 1/9x^-2 \ dx$
$" "= x^3/3 + 1/9(-1)x^-1 +c$
$" "= x^3/3 - 1/(9x) +c$

How do you integrate by substitution int [x^2+1/(3x)^2]dx∫[x2+1(3x)2]dxint [x^2+1/(3x)^2]dx?