Evaluate the following integrals. ∫81x3−9x2dx

Question

Evaluate the following integrals.  ∫81x3−9x2dx

Joseph Fair · Accepted Answer

Step 1  Given: I=∫81x3−9x2dx  for evaluating given integral, we first simplify it then integrate is  so,  ∫81dxx3−9x2=81∫dxx2(x−9)  =81∫[−181x−19x2+181(x−9)]dx  =81[−181ln|x|+19x+181ln|x−9|]+c  (∵∫dxx−a=ln|x−a|+c,∫xndx=xn+1n+1+c  =−ln|x|+9x+ln|x−9|+c)  Step 2  hence, given integral is ln|x−9|−ln|x|+9x+c.

redhotdevil13l3 · Accepted Answer

∫81x3−9x2dx  81⋅∫1x3−9x2dx  81⋅∫−181x−19x2+181(x−9)dx  81(−∫181xdx−∫19x2+181(x−9))dx  81(−181⋅ln⁡(|x|)+19x+181⋅ln⁡(|x−9|))  −ln⁡(|x|)+9x+ln⁡(|x−9|)  Solution:  −ln⁡(|x|)+9x+ln⁡(|x−9|)+C

Vasquez · Accepted Answer

∫81x3−9∗x2dx Let&#039;s represent it in the form: 81x2∗(x−9)=81x2∗(x−9) We use the method of decomposition into the elementary elements. Let us expand the function into the simplest terms: 81x2(x−9)=Ax+Bx2+Cx−9=Ax(x−9)+B(x−9)+Cx2x2(x−9) 81=Ax(x−9)+B(x−9)+Cx2 x2:A+C=0 x:-9A+B=0 1: -9B=81 Solving it, we find: A=-1; B=-9; C=1 81x2(x−9)=−1x+−9x2+1x−9 We calculate the tabular integral: We ∫1x−9dx=ln⁡(x−9) calculate the tabular integral: We −∫1xdx=−ln⁡(x) calculate the tabular integral: ∫−9x2dx=9x Answer: ln⁡(x−9)−ln⁡(x)+9x+C

Evaluate the following integrals. \int \frac{81}{x^{3}-9x^{2}}dx

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