Integrate the function: \(\frac{6x+7}{√(x-5)(x-4)}\)
Solution and Explanation
\(\frac{6x+7}{√(x-5)(x-4)}\) =\(\frac{6x+7}{√x^2-9x+20}\)
Let 6x+7 = A\(\frac{d}{dx}\)(x2-9x+20)+B
⇒ 6x+7 = A(2x-9)+B
Equating the coefficients of x and constant term, we obtain
2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34
∫\(\frac{6x+7}{√x^2-9x+20}\) = ∫\(\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}\) dx
=3 ∫\(\frac{(2x-9)}{\sqrt{x2-9x+20}}\)+ 34 ∫\(\frac{1}{\sqrt{x2-9x+20}}\) dx
Let I1 = ∫\(\frac{(2x-9)}{\sqrt{x^2-9x+20}}\) and I2 = ∫\(\frac{1}{\sqrt{x^2-9x+20}}\)dx
∴ ∫\(\frac{(6x+7)}{\sqrt{x^2-9x+20}}\) = 3I1+34I2 ...(1)
Then,
I1 = ∫\(\frac{(2x-9)}{\sqrt{x^2-9x+20}}\) dx
Let x2-9x+20 = t
⇒ (2x-9)dx = dt
⇒I1 = dt/√t
I1 = 2√t
I1 = 2√x2-9x+20 ...(2)
and I2 = ∫\(\frac{1}{\sqrt{x^2-9x+20}}\)dx
x2-9x+20 can be written as x2-9x+20+\(\frac{81}{4}\)-\(\frac{81}{4}\)
Therefore,
x2-9x+20+\(\frac{81}{4}\)-\(\frac{81}{4}\)
=(x-\(\frac{9}{2}\))2-1/4
=(x-\(\frac{9}{2}\))2-(1/2)2
⇒ I2 = ∫1/√(x-\(\frac{9}{2}\))2-(1/2)2 dx
I2 = log|(x-\(\frac{9}{2}\))+√x2-9x+20| ...(3)
Substituting equations (2) and (3) in (1), we obtain
∫6x+7/√x2-9x+20 dx = 3[2√x2-9x+20]+34 log[(x-\(\frac{9}{2}\))+√x2-9x+20]+C
=6√x2-9x+20 + 34 log[(x-\(\frac{9}{2}\))+√x2-9x+20]+C
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- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
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- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
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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.
