Question

Please help with each box of the problem.

EXAMPLE 5 Evaluate 7x2 - 6x + 16 x3 + 4x dx. Since x3 + 4x = x(x2 + 4) can't be factored further, we SOLUTION write А. Bx + C 7x² - 6x + 16 x(x2 + 4) + Multiplying by x(x2 + 4), we have 7x2 - 6x + 16 = A(x2 + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 7 C = -6 4A = 16. Thus A = B = and C= -6 and so 7x² - 6x + 16 dx = SC + X - 6 x² + 4 dx. x3 + 4x х In order to integrate the second term we split it into two parts: x2 X-6 6 dx = dx - dx. +4 x² + 4 x2 + 4 We make the substitution u = x2 + 4 in the first of these integrals so that du = 2x dx. We evaluate the second integral by means of this formula with a = 2. (Remember to use absolute values where appropriate.) 7x2 - 6x + 16 x(x2 + 4) dx = dx + 6 x² + 4 +K

Answer 1

7x² - 6x +16 doc x3tha + Bx+C x 1x²+4 + CathA. = Evaluate Ans-> 7x²60c +16 А a 192+) multiplying by x(22+4), we have, 7X²-6x+16= A (x²+4) + (Box+ ( a Ax² + Boc? +Cx+hA |X²(A+B) - Equating co-efficients, we obtain, A+B=7, (A = 7-A, -7-6-3) thing A-G B Ja sam hteo sa Bana Jose an orded to integrate the second toom 10x260x53. doc - Jobs make the substitution re=x2+4 C=-6 ,4A = 16 ) 3 & C=-6 and so split it into parts doc xen du=2x da,

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