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Question
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
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Solution
Let I = `int (3x - 1)/sqrt(x^2 + 9) "d"x`
= `int (3x)/sqrt(x^2 + 9) "d"x - int 1/sqrt(x^2 + 9) "d"x`
I = I1 – I2
Now I1 = `int (3x)/sqrt(x^2 + 9) "d"x`
Put x2 + 9 = t
⇒ 2x dx = dt
x dx = – dt
∴ I1 = `3/2 int "dt"/sqrt("t")`
= `3/2 * 2sqrt("t") + "C"_1`
= `3sqrt(x^2 + 9) + "C"_1`
I2 = `int 1/sqrt(x^2 + 9) "d"x`
= `int 1/sqrt(x^2 + (3)^2) "d"x`
= `log|x + sqrt(x^2 + (3)^2)| + "C"_2` ....`[because int 1/sqrt(x^2 + "a"^2) "d"x = log|x + sqrt(x^2 + "a"^2)| + "C"]`
= `log|x + sqrt(x^2 + 9)| + "C"_2`
∴ I = I1 – I2
= `3sqrt(x^2 + 9) + "C"_1 - log|x + sqrt(x^2 + 9)| - "C"_2`
= `3sqrt(x^2 + 9) - log|x + sqrt(x^2 + 9)| + ("C"_1 - "C"_2)`
Hence, I = `3sqrt(x^2 + 9) - log|x + sqrt(x^2 + 9)| + "C"`
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