How do we integrate : ∫▒〖x tan^(-1)⁡3x 〗 dx?

Taking from where you left off:

∫ 3 / (1 + (9x)²) * x²/2 dx

= 3/2 ∫ x² / (1 + (9x)²) dx

= 3/2*1/9 ∫ 9x² / (1 + (9x)²) dx

= 3/18 ∫ (1+9x² - 1) / (1 + (9x)²) dx

= 3/18 ∫ 1 - 1 / (1 + (9x)²) dx

= 3x/18 - 3/18∫1 / (1 + (9x)²) dx

= 3x/18 - 1/54 tan⁻¹9x + C

∫ 3/(1+ 〖9x〗^2 ) * x^2/2 dx

Now, in this part , you can take 3/2 outside the integration and you are left with x^2/ 9x^2 +1

SO the power of x in denominator and nominator is same, so you need to do long division.

So you get 1/9 ∫ dx - 1/9 ∫ (9x^2 +1 ) dx

x/9 - 1/9 ∫ ( 9x^2) dx - 1/9 ∫ dx

x/9 - x^3 / 3 - x/9

- x^3/ 3

? dx/(x^2 + 5) = (a million/5) ? dx/(x^2/5 + a million) enable x = ?5 tan t dx = ?5 sec^2 t dt (a million/5) ? dx/(x^2/5 + a million) = (a million/5) ? ?5 sec^2t dt/(tan^2 t + a million) = (a million/?5) ? sec^2 t dt/sec^2 t = (a million/?5) ? dt = (a million/?5) t + C = a million/?5 arctan (x/?5) + C