Question

Evaluate the following:

`int (3x - 1)/sqrt(x^2 + 9) "d"x`

Answer 1

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"`