Evaluate the integral. ∫1x(x−a)dx

Question

Evaluate the integral.  ∫1x(x−a)dx

psor32 · Accepted Answer

Step 1  Given data:  The integral expression is: ∫1x(x−a)dx  The given integral can be expressed as,  ∫1x(x−a)dx=∫1ax(xa−1)dx  =1a∫1x(xa−1)dx  Assume x/a =u, and differentiate it with respect to x.  1adx=du  dx=adu  Step 2  Substitute the above-calculated values in the expression.  ∫1x(x−a)dx=1a∫1(au)(u−1)(adu)  =1a∫duu(u−1)  =1a∫(−1u+1u−1)du  =1a(−ln|u|+ln|u−1|)+C  =1aln|u−1u|+C  Substitute x/a for u in the expression.  ∫1x(x−a)dx=1aln|(xa)−1(xa)|+C  =1aln|x−ax|+C  Thus, the integration of the given expression is ln|x−ax|a+C.

Dawn Neal · Accepted Answer

Step 1  We can write  1x(x−a)=Ax+Bx−a  Multiply both sides by x(x-a)  1=A(x-a)+Bx  1=Ax-Aa+Bx  Group the like terms  0*x+1=(A+B)x-Aa  1=-Aa  Divide both sides by -a  −1a=A  0=A+B  Substitute A with −1a, to get  0=−1a+B  Add 1a to both sides  1a=B  Step 2  Substitute back the values of A and B, to get  ∫1x(x−a)dx=∫1ax−1ax−adx  =1aln|x|−1aln|x−a|+C  =1aln|xx−a|+C  Result:  ∫1x(x−a)dx=1aln|xx−a|+C

karton · Accepted Answer

Given:∫1x(x−a)dx=∫1(1−ax)x2dx=1a∫1udu∫ln⁡(u)a=ln⁡(1−ax)a=ln⁡(|ax−1|)a+Csimplify:ln⁡(|x−a|)−ln⁡(|x|)a+C (Answer)

Evaluate the integral. \int \frac{1}{x(x-a)}dx

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