Question

# Integrate the following functions. ∫6x+7√(x−5)(x−4)dx

Solution

## ∫6x+7√(x−5)(x−4)dx=∫6x+7√x2−9x+20dx Let 6x+7=Addx(x2−9x+20)+B ⇒6x+7=A(2x−9)+B⇒6x+7=2Ax+(−9A+B) On equating the coefficients of x and constant term on both sides, we get 2A=6⇒A=3 and −9A+B=7⇒B=34 ⇒∫6x+7√x2−9x+20dx=∫3(2x−9)+34√x2−9x+20dx=3∫2x−9√x2−9x+20dx+34∫1√x2−9x+20dx Let I1=∫2x−9√x2−9x+20dx and I2=∫1√x2−9x+20dx ∴∫6x+7√x2−9x+20dx=3I1+34I2......(i) Now, I1=∫2x−9√x2−9x+20dx Let x2−9x+20=t⇒(2x−9)dx=dt ∴I1=∫dt√t=2√t+C1=2√x2−9x+20+C1....(ii) And I2=∫1√x2−9x+20dx x2−9x+20can be written as x2−9x+20+814−814 Therefore, x2−9x+20+814−814=(x−92)2−14=(x−92)2−(12)2 ∴I2=∫1√(x−92)2−(12)2dx=log∣∣∣(x−92)+√(x−92)2−14∣∣∣+C2 [∵∫dx√x2−a2=log|x+√x2−a2|]=log∣∣(x−92)+√x2−9x+20∣∣+C2......(iii) On substituting the values of I1 and I2 from Eqs.(ii)and (iii)in Eq. (i). we get ∫6x+7√x2−9x+20dx=3[2√x2−9x+20]+34log∣∣(x−92)+√x2−9x+20∣∣+C[∵3C1+34C2=C]=6√x2−9x+20+34log∣∣(x−92)+√x2−9x+20∣∣+C

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