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QuestionFind the integral of 1−x2x(1−2x)\frac{1-x^{2}}{x(1-2 x)} and simplify to get the final expression.

Studdy Solution

STEP 1

Assumptions1. The integral is a proper integral and can be solved using the method of partial fractions. . The function is continuous over the interval of integration.

STEP 2

The given integral is∫1−x2x(1−2x)dx\int \frac{1-x^{2}}{x(1-2 x)} d xWe can rewrite this as∫(12+(−x2+1)x(1−2x))dx\int\left(\frac{1}{2}+\frac{\left(-\frac{x}{2}+1\right)}{x(1-2 x)}\right) d x

STEP 3

Now, we can split the integral into two parts as follows=12∫1dx+∫−x2+1x(1−2x)dx=\frac{1}{2} \int1 d x+\int \frac{-\frac{x}{2}+1}{x(1-2 x)} d x

STEP 4

Let's assume that the integrand −x2+1x(1−2x)\frac{-\frac{x}{2}+1}{x(1-2 x)} can be written as Ax+B1−2x\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-2 x}.

STEP 5

Multiplying by the least common multiple (LCM), we get−x2+1=A(1−2x)+Bx=A−2 Ax+Bx-\frac{x}{2}+1=\mathrm{A}(1-2 x)+\mathrm{B} x=\mathrm{A}-2 \mathrm{~A} x+\mathrm{B} x

STEP 6

Comparing the constants, we find that A=1\mathrm{A}=1.

STEP 7

Substituting A=1\mathrm{A}=1 into the equation, we get−2+B=−12⇒B=−12+2=−1+42-2+B=\frac{-1}{2} \Rightarrow B=\frac{-1}{2}+2=\frac{-1+4}{2}So, B=32B=\frac{3}{2}.

STEP 8

Substituting the values of A\mathrm{A} and B\mathrm{B} into the equation, we get−x2+1x(1−2x)=1x+321−2x\frac{-\frac{x}{2}+1}{x(1-2 x)} =\frac{1}{x}+\frac{\frac{3}{2}}{1-2 x}

STEP 9

Now, we can integrate the above equation∫−x2+x(−2x)dx=∫xdx+32∫−2xdx\int \frac{-\frac{x}{2}+}{x(-2 x)} d x =\int \frac{}{x} d x+\frac{3}{2} \int \frac{}{-2 x} d x

STEP 10

olving the integrals, we get=log⁡∣x∣+32log⁡∣−2x∣−2+c= \log |x|+\frac{3}{2} \log \frac{|-2 x|}{-2}+c

STEP 11

implifying the above equation, we get=log⁡∣x∣−34log⁡∣−x∣+c= \log |x|-\frac{3}{4} \log |- x|+c

STEP 12

Substituting this value back into the original integral, we get the final solution∫−x2x(−2x)dx=2x+log⁡∣x∣−4log⁡∣−2x∣+c\int \frac{-x^{2}}{x(-2 x)} d x=\frac{}{2} x+\log |x|-\frac{}{4} \log |-2 x|+c

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