If (x+2) {x²-x+2} dx = \;1/3 f(x) + \;5/8 g(x) + \;35/16 h(x) + C then f(-1) + g(-1) + h≤ft(1/2) =

Correct Answer is : 2+ln⁡(72)2 + ≤ft( {{7}}{2} )2+ln(27)

Correct Answer is : 2+ln⁡(72)2 + ≤ft( {{7}}{2} )2+ln(27)

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TG EAMCET 2 May 2025 Shift 1 Paper

Question : 75 of 160

Marks: +1, -0

\int (x+2) \sqrt{x^2-x+2} dx = \;\frac{1}{3} f(x) + \;\frac{5}{8} g(x) + \;\frac{35}{16} h(x) + C

f(-1) + g(-1) + h\left(\frac{1}{2}\right) =

Solution:

Given,

\int (x+2) \sqrt{x^2-x+2} dx

\Rightarrow\;\; \;\frac{1}{3} f(x) + \;\frac{5}{8} g(x) + \;\frac{35}{16} h(x) + C

Let

x+2 = A(2x-1) + B

Comparing coefficients, we get

2A = 1

\Rightarrow A = \frac{1}{2}, -A+B = 2

\Rightarrow B = 2+A = 2+\frac{1}{2} = \frac{5}{2}

So,

x+2 = \frac{1}{2}(2x-1) + \frac{5}{2}

\int (x+2) \sqrt{x^2-x+2} dx

= \int \left( \frac{1}{2}(2x-1) + \frac{5}{2} \right) \sqrt{x^2-x+2} dx

= \frac{1}{2} \int (2x-1) \sqrt{x^2-x+2} dx + \frac{5}{2} \int \sqrt{x^2-x+2} dx

Let

u = x^2 - x + 2

, then

du = (2x-1) dx

= \frac{1}{2} \int \sqrt{u} du + \frac{5}{2} \int \sqrt{ \left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2 } dx

= \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + \frac{5}{2} \left[ \frac{2x-1}{4} \sqrt{x^2-x+2} + \frac{7}{8} \ln \left| x-\frac{1}{2}+\sqrt{x^2-x+2} \right| \right] + c

= \frac{1}{3} (x^2-x+2)^{3/2} + \frac{5(2x-1)}{8} \sqrt{x^2-x+2} + \frac{35}{16} \ln \left| x-\frac{1}{2}+\sqrt{x^2-x+2} \right| + c

Comparing with original equation, we get

f(x) = (x^2-x+2)^{3/2}

g(x) = (2x-1) \sqrt{x^2-x+2}

h(x) = \ln \left| x-\frac{1}{2}+\sqrt{x^2-x+2} \right|

So,

f(-1) = [(-1)^2 - (-1) + 2]^{3/2}

= [1+1+2]^{3/2} = 4^{3/2} = 8

g(-1) = (2(-1)-1) \sqrt{(-1)^2 - (-1) + 2}

= -3 \sqrt{1+1+2} = -3 \sqrt{4}

= -3 \times 2 = -6

h\left(\frac{1}{2}\right) = \ln \left| \frac{1}{2} - \frac{1}{2} + \sqrt{ \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 2 } \right|

\Rightarrow \ln \left| 0 + \sqrt{ \frac{1}{4} - \frac{1}{2} + 2 } \right|

\Rightarrow \ln \left| \sqrt{ \frac{1-2+8}{4} } \right| \Rightarrow \ln \left| \sqrt{ \frac{7}{4} } \right| = \ln \left( \frac{\sqrt{7}}{2} \right)

\;\;\text{So, }\; f(-1) + g(-1) + h\left(\frac{1}{2}\right) = 8 - 6 + \ln \left( \frac{\sqrt{7}}{2} \right)

\;= 2 + \ln \left( \frac{\sqrt{7}}{2} \right)

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