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Question Number 136905 by leena12345 last updated on 27/Mar/21

∫_0 ^3 ((56)/(x^2 −6x+5))dx

∫3056x2−6x+5dx

Answered by Mathspace last updated on 27/Mar/21

Ψ=∫_0 ^3 ((56)/(x^2 −6x+5)) Δ^′ =(−3)^2 −5=4 ⇒x_1 =3+2=5 x_2 =3−2=1 ⇒ Ψ=56∫ (dx/((x−1)(x−5))) =((56)/4)∫((1/(x−5))−(1/(x−1)))dx =14ln∣((x−5)/(x−1))∣ +C

Ψ=∫0356x2−6x+5Δ′=(−3)2−5=4⇒x1=3+2=5x2=3−2=1⇒Ψ=56∫dx(x−1)(x−5)=564∫(1x−5−1x−1)dx=14ln∣x−5x−1∣+C

Commented by Mathspace last updated on 27/Mar/21

Ψ=14[ln∣((x−5)/(x−1))∣]_0 ^3 =14{ln(1)−ln(5)}=−14ln(5)

Ψ=14[ln∣x−5x−1∣]03=14{ln(1)−ln(5)}=−14ln(5)

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