Evaluate-1-ax-2-bx-c-dx-

Question Number 83807 by niroj last updated on 06/Mar/20

Evaluate :

\int \frac{1}{{ax}^{2} + bx + c} dx

Commented by mathmax by abdo last updated on 06/Mar/20

I = \int \frac{d x}{{a x}^{2} + b x + c}

c a s e 1 a = 0 a n d b \neq 0 \Rightarrow I = \frac{1}{b} l n ∣ b x + c ∣ + K n o w w e t a k e a \neq 0

Δ = b^{2} - 4 a c c a s e 1 Δ > 0 \Rightarrow x_{1} = \frac{- b + \sqrt{b^{2} - a c}}{2 a} a n d x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}

I = \int \frac{d x}{a (x - x_{1}) (x - x_{2})} = \frac{1}{a} \int (\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}) d x

= \frac{1}{\sqrt{b^{2} - 4 a c}} l n ∣ \frac{x - x_{1}}{x - x_{2}} ∣ + C

c a s e 2 Δ < 0 \Rightarrow b^{2} < 4 a c \Rightarrow I = \int \frac{d x}{a (x^{2} + \frac{b}{a} x + \frac{c}{a})}

= \frac{1}{a} \int \frac{d x}{(x^{2} + 2 \frac{b}{2 a} x + \frac{b^{2}}{4 a^{2}} + \frac{c}{a} - \frac{b^{2}}{4 a^{2}})}

= \frac{1}{a} \int \frac{d x}{{(x + \frac{b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a^{2}}} =_{x + \frac{b}{2 a} = \sqrt{\frac{4 a c - b^{2}}{4 a^{2}}} z} \frac{1}{a} \int \frac{1}{\frac{4 a c - b^{2}}{4 a^{2}} (1 + z^{2})} \times \sqrt{\frac{4 a c - b^{2}}{4 a^{2}}} d z

= \frac{4 a}{4 a c - b^{2}} \times \frac{\sqrt{4 a c - b^{2}}}{2 ∣ a ∣} a r c t a n (\frac{1}{\sqrt{\frac{4 a c - b^{2}}{4 a^{2}}}} (x + \frac{b}{2 a})) + K

= \frac{2 a}{∣ a ∣ \sqrt{4 a c - b^{2}}} a r c t a n (\frac{1}{\sqrt{\frac{4 a c - b^{2}}{4 a^{2}}}} (x + \frac{b}{2 a})) + K

=

Answered by john santu last updated on 06/Mar/20

\frac{1}{a} \int \frac{dx}{x^{2} + \frac{b}{a} x + \frac{c}{a}} =

\frac{1}{a} \int \frac{dx}{{(x + \frac{b}{2 a})}^{2} + \frac{c}{a} - \frac{b^{2}}{{4 a}^{2}}} =

\frac{1}{a} \int \frac{dx}{{(x + \frac{b}{2 a})}^{2} + {(\sqrt{\frac{4 ac - b^{2}}{{4 a}^{2}}})}^{2}} =

\frac{2 a}{\sqrt{4 ac - b^{2}}} \tan^{- 1} (\frac{{4 a}^{2} x + 2 ab}{\sqrt{4 ac - b^{2}}}) + c

Answered by TANMAY PANACEA last updated on 06/Mar/20

{a x}^{2} + b x + c

\frac{1}{4 a} (4 a^{2} x^{2} + 4 a b x + 4 a c)

\frac{1}{4 a} {{(2 a x + b)}^{2} + 4 a c - b^{2}}

c o n d i t i o n \dots (1)

w h e n b^{2} > 4 a c

\frac{1}{4 a} {{(2 a x + b)}^{2} - (b^{2} - 4 a c)}

\frac{1}{4 a} {{(2 a x + b)}^{2} - {(\sqrt{b^{2} - 4 a c})}^{2}}

s o \int \frac{d x}{{a x}^{2} + b x + c}

\int \frac{d x}{\frac{1}{4 a} {{(2 a x + b)}^{2} - {(\sqrt{b^{2} - 4 a c})}^{2}}

n o w p l s u s e f o r m u l a \int \frac{d y}{y^{2} - k^{2}}

c o n d i t i o n 2

b^{2} < 4 a c

\int \frac{d x}{\frac{1}{4 a} {{(2 a x + b)}^{2} + {(\sqrt{4 a c - b^{2}})}^{2}}

n o w u s e f o r m u l a \int \frac{d y}{y^{2} + k^{2}}

c o n d i t i o n 3

b^{2} = 4 a c

\int \frac{d x}{\frac{1}{4 a} {(2 a x + b)}^{2}}

Commented by niroj last updated on 06/Mar/20

t h a n k s t o a l l .

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