BeMath-x-2-ln-x-2-3-dx-

Question Number 107756 by bemath last updated on 12/Aug/20

\frac{B e M a t h}{∐}

\int x^{2} \ln (x^{2} + 3) d x

Answered by hgrocks last updated on 12/Aug/20

I = \ln (x^{2} + 3) \frac{x^{3}}{3} - \frac{2}{3} \int \frac{x^{4}}{x^{2} + 3} dx \to J

J = \int \frac{x^{4} - 9 + 9}{x^{2} + 3} dx

= \int (x^{2} - 3) dx + 9 \int \frac{dx}{x^{2} + 3}

= \frac{x^{3}}{3} - 3 x + \frac{9}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}})

So I = \frac{x^{3}}{9} (3 \ln (x^{2} + 3) - 2) + 2 (x - \sqrt{3} \tan^{- 1} (\frac{x}{\sqrt{3}}))

Answered by bobhans last updated on 12/Aug/20

\frac{⊺ B obhans ⊺}{Π}

set \ln (x^{2} + 3) = u \Rightarrow \frac{2 x}{x^{2} + 3} dx = du

v = \frac{1}{3} x^{3}

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{1}{3} \int \frac{{2 x}^{4}}{(x^{2} + 3)} dx

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{2}{3} \int \frac{x^{4}}{x^{2} + 3} dx

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{2}{3} \int \frac{{(x^{2} + 3)}^{2} - {6 x}^{2} - 9}{x^{2} + 3} dx

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{2}{3} \int ((x^{2} + 3) - \frac{6 (x^{2} + 3) + 9}{x^{2} + 3}) dx

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{2}{3} [\frac{1}{3} x^{3} + 3 x - 6 x + \int \frac{9}{x^{2} + 3} dx]

I = \frac{1}{3} x^{3} \ln (x^{2} + 3) - \frac{2}{9} x^{3} + 2 x - 2 \sqrt{3} \tan^{- 1} (\frac{x}{\sqrt{3}}) + C

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