Question:

Integrate the function: \(\frac {1}{ {9x^2+6x+5}}\)

Updated On: Oct 4, 2023
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Solution and Explanation

\(∫\)\(\frac {1}{9x^2+6x+5} dx\) = \(∫\frac {1}{(3x+1)^2+(2)^2}dx\)

\(Let \ (3x+1) = t\)

\(∴ 3dx = dt\)

⇒ \(∫\frac {1}{(3x+1)^2+(2)^2}dx\) \(=\) \(\frac 13 ∫\frac {1}{t^2+2^2} dt\)

                                           \(=\frac 13 [\frac 12\ tan^{-1}(\frac t2)]+C\)

                                           \(=\frac 16\ tan^{-1}(\frac {3x+1}{2})+C\)

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Concepts Used:

Integrals of Some Particular Functions

There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.

Integrals of Some Particular Functions:

  • ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
  • ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
  • ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
  • ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
  • ∫1/√(a2 – x2) dx = sin-1(x/a) + C
  • ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C

These are tabulated below along with the meaning of each part.