261.
261.
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1 Answer
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Let
The integrand is of the form.
1 = Ax (x + 1) + B (x + 1) + Cx2
= A (x2 + x) + B (x + 1) + Cx2
Comparing the coefficients,
A + C = 0 ____ (1)
A + B = 0 ______ (2)
B = 1 ________ (3)
Putting Equation (3) in (2),
A + 1 = 0
A = -1.
and putting value of A in Equation (1),
-1 + C = 0
C = 1
Hence proved.
Similar Questions for you
Let 2x = t then
Let tan t = y then
Let
The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
This is a complex integral. The provided solution splits it into two parts:
I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
10 (α + βγ + δ) = 10 (1 + (1/2√5)*√5 + 1/2) seems incorrect.
The calculation is shown as 10 (1
The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.
The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)
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