Question-97775

Question Number 97775 by liki last updated on 09/Jun/20

Commented by liki last updated on 09/Jun/20

. . i need help

Commented by Tony Lin last updated on 09/Jun/20

\int \frac{x^{2}}{x^{2} + b x + 12} d x

= \int \frac{x^{2} + b x + 12}{x^{2} + b x + 12} - \frac{b}{2} \int \frac{2 x + b}{x^{2} + b x + 12} d x

- \frac{b}{2} \int \frac{\frac{24}{b} - b}{{(x + \frac{b}{2})}^{2} + {(\sqrt{12 - \frac{b^{2}}{4}})}^{2}} d x

= x - \frac{b}{2} l n ∣ x^{2} + b x + 12 ∣ + \frac{b^{2} - 24}{\sqrt{48 - b^{2}}} {t a n}^{- 1} \frac{2 x + b}{\sqrt{48 - b^{2}}} + c

Answered by Ar Brandon last updated on 09/Jun/20

I = \int \frac{x^{2}}{x^{2} + bx + 12} dx = \int {1 - \frac{bx + 12}{x^{2} + bx + 12}}

= \int {1 - \frac{b}{2} \cdot \frac{2 x + b}{x^{2} + bx + 12} + \frac{\frac{b^{2}}{2} - 12}{x^{2} + bx + 12}} dx

= \int {1 - \frac{b}{2} \cdot \frac{2 x + b}{x^{2} + bx + 12} + 2 \cdot \frac{b^{2} - 24}{{(2 x + b)}^{2} + (48 - b^{2})}} dx

= x - \frac{b}{2} \cdot \ln ∣ x^{2} + bx + 12 ∣ + \frac{b^{2} - 24}{\sqrt{48 - b^{2}}} \tan^{- 1} [\frac{2 x + b}{\sqrt{48 - b^{2}}}] + C

Answered by abdomathmax last updated on 09/Jun/20

I = \int \frac{x^{2}}{x^{2} + bx + 12} dx \Rightarrow I = \int \frac{x^{2} + bx + 12 - bx - 12}{x^{2} + bx + 12} dx

= x - \int \frac{bx + 12}{x^{2} + bx + 12} dx

x^{2} + bx + 12 = 0 \to Δ = b^{2} - 48

case 1 b^{2} - 48 > 0 \Rightarrow x_{1} = \frac{- b + \sqrt{b^{2} - 48}}{2}

x_{2} = \frac{- b - \sqrt{b^{2} - 48}}{2}

f (x) = \frac{bx + 12}{x^{2} + bx + 12} = \frac{bx + 12}{(x - x_{1}) (x - x_{2})} = \frac{a}{x - x_{1}} + \frac{b}{x - x_{2}}

a = \frac{{bx}_{1} + 12}{\sqrt{b^{2} - 48}} and b = - \frac{{bx}_{2} + 12}{\sqrt{b^{2} - 48}}

\int f (x) dx = aln ∣ x - x_{1} ∣ + bln ∣ x - x_{2} ∣ + c \Rightarrow

I = x - aln ∣ x - x_{1} ∣ - bln ∣ x - x_{2} ∣ + C

case 2 b^{2} - 48 < 0 \Rightarrow \frac{bx + 12}{x^{2} + bx + 12}

= \frac{bx + 12}{x^{2} + 2 x \frac{b}{2} + \frac{b^{2}}{4} + 12 - \frac{b^{2}}{4}} = \frac{bx + 12}{{(x + \frac{b}{2})}^{2} + \frac{48 - b^{2}}{4}}

we do the changement x + \frac{b}{2} = \frac{\sqrt{48 - b^{2}}}{2} u \Rightarrow

\int \frac{bx + 12}{x^{2} + bx + 12} dx = \frac{4}{\sqrt{48 - b^{2}}} \int \frac{b (\frac{\sqrt{48 - b^{2}}}{2} u - \frac{b}{2}) + 12}{1 + u^{2}} \times \frac{\sqrt{48 - b^{2}}}{2} du

= 2 \int \frac{\sqrt{48 - b^{2}} u - \frac{b^{2}}{2} + 12}{1 + u^{2}} du

= \sqrt{48 - b^{2}} \ln (1 + u^{2}) + (24 - b^{2}) \arctan (u) + C

= \sqrt{48 - b^{2}} \ln (1 + {(\frac{2 x + b}{\sqrt{48 - b^{2}}})}^{2})

+ (24 - b^{2}) \arctan (\frac{2 x + b}{\sqrt{48 - b^{2}}}) + C \Rightarrow

I = x - \sqrt{48 - b^{2}} \ln (1 + \frac{{(2 x + b)}^{2}}{48 - b^{2}})

- (24 - b^{2}) \arctan (\frac{2 x + b}{\sqrt{48 - b^{2}}}) + C

Answered by niroj last updated on 09/Jun/20

\int \frac{x^{2}}{x^{2} + bx + 12} dx

= \int \frac{x^{2} + bx + 12 - bx - 12}{x^{2} + bx + 12} dx

= \int \frac{x^{2} + bx + 12}{x^{2} + bx + 12} dx - \int \frac{bx + 12}{x^{2} + bx + 12} dx

= \int dx - \int \frac{\frac{b}{2} (2 x + b) + 12 - \frac{b^{2}}{2}}{x^{2} + bx + 12} dx

= \int dx - \frac{b}{2} \int \frac{(2 x + b)}{x^{2} + bx + 12} dx - (12 - \frac{b^{2}}{2}) \int \frac{1}{x^{2} + bx + 12} dx

= \int dx - \frac{b}{2} \int \frac{(2 x + b)}{x^{2} + bx + 12} dx - (12 - \frac{b^{2}}{2}) \int \frac{1}{{(x)}^{2} + 2 x . \frac{b}{2} + \frac{b^{2}}{4} - \frac{b^{2}}{4} + 12} dx

= \int dx - \frac{b}{2} \int \frac{(2 x + b) dx}{x^{2} + bx + 12} - (12 - \frac{b^{2}}{2}) \int \frac{1}{{(x + \frac{b}{2})}^{2} - (\frac{b^{2} - 48}{4})} dx

= \int dx - \frac{b}{2} \int \frac{(2 x + b) dx}{x^{2} + bx + 12} - (12 - \frac{b^{2}}{2}) \int \frac{1}{{(x + \frac{b}{2})}^{2} - {(\frac{\sqrt{b^{2} - 48}}{2})}^{2}} dx

= x - \frac{b}{2} \log (x^{2} + bx + 12) - (12 - \frac{b^{2}}{2}) . \frac{1}{2 . \frac{\sqrt{b^{2} - 48}}{2}} \log \frac{x + \frac{b}{2} - \frac{\sqrt{b^{2} - 48}}{2}}{x + \frac{b}{2} + \frac{\sqrt{b^{2} - 48}}{2}} + C

= x - \frac{b}{2} \log (x^{2} + bx + 12) - (12 - \frac{b^{2}}{2}) \frac{1}{\sqrt{b^{2} - 48}} \log \frac{2 x - b - \sqrt{b^{2} - 48}}{2 x + b + \sqrt{b^{2} - 48}} + C / / .

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