Best solutions to - "Find the indefinite integral. ∫1x2−4x+9dx"

Michelle Arroyo

Answered question

2021-11-07

Find the indefinite integral.

\int \frac{1}{x^{2} - 4 x + 9} d x

Charles Randolph

Beginner2021-11-08Added 16 answers

Step 1
the given integral is:

\int \frac{1}{x^{2} - 4 x + 9} d x

we have to find the value of the given integral.
let the given integral be I.
therefore,

I = \int \frac{1}{x^{2} - 4 x + 9} d x

Step 2

I = \int \frac{1}{x^{2} - 4 x + 9} d x

= \int \frac{1}{x^{2} - 4 x + 2^{2} - 2^{2} + 9} d x

= \int \frac{1}{x^{2} - 4 x + 2^{2} + 9 - 4} d x

= \int \frac{1}{{(x - 2)}^{2} + {(\sqrt{5})}^{2}} d x

as we know that

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

Step 3
therefore,

I = \int \frac{1}{{(x - 2)}^{2} + {(\sqrt{5})}^{2}} d x

= \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{x - 2}{\sqrt{5}}) + C

where C is the constant of integration.
therefore,

I = \int \frac{1}{x^{2} - 4 x + 9} d x = \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{x - 2}{\sqrt{5}}) + C

therefore the value of the given integral

\int \frac{1}{x^{2} - 4 x + 9} d x i s \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{x - 2}{\sqrt{5}}) + C

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