Anti-differentiate reciprocal and linear-reciprocal functions

Which rule applies first to the reciprocal integrand?

Derivative and boundary condition

\(\displaystyle F^{\prime}(x)=\frac{8}{2x+3},\quad x>-\frac32,\quad F(u)=7\)

Change the boundary input u. Compare the constant with the unchanged condition F(u)=7

Why does the additive constant change?

Anti-differentiate reciprocal functions

The reciprocal of a linear expression has a logarithmic anti-derivative. Divide the numerator coefficient by the coefficient inside the denominator, use an absolute value on a general branch and include C for an indefinite integral

1. 1Divide by the inner coefficient Divide the numerator coefficient by the coefficient of x in the denominator \(\displaystyle \int\frac{6}{2x+1}\,dx=3\log_e\left|2x+1\right|+C\)
2. 2Use an absolute value on a full branch The absolute value keeps the logarithm argument positive on either non-zero branch \(\displaystyle \int\frac{1}{x}\,dx=\log_e\left|x\right|+C\)
3. 3Keep one arbitrary constant An indefinite integral represents a family of anti-derivatives, so include one arbitrary constant \(\displaystyle \int\left(f(x)+g(x)\right)\,dx=F(x)+G(x)+C\)
4. 4Combine terminal logarithms For a definite reciprocal integral, combine the endpoint logarithms into one positive ratio \(\displaystyle \int_1^4\frac{2}{x+1}\,dx=2\log_e\left(\frac{5}{2}\right)\)

Reciprocal integral

\(\displaystyle \int 6/(2x+1)\,dx\)

Find the coefficient outside the logarithm for the displayed reciprocal integral

Reciprocal integral

\(\displaystyle \int_1^4 6/(2x+1)\,dx\)

Find the positive logarithm argument by taking the upper inner value over the lower inner value

Missing the reciprocal inner coefficient. One of these lines is right. Which one?

A non-unit argument leaves a logarithmic constant

On x>−3/2, integrate the derivative first: F(x)=4 ln(2x+3)+C. Substitute the complete condition F(1)=7: 4 ln 5+C=7, so C=7−4 ln 5, not 7. A single-log form is F(x)=7+4 ln((2x+3)/5). The argument is positive on the specified connected domain. Differentiating gives 8/(2x+3), and substitution at 1 gives 7+4 ln 1=7. Both checks are required

Construct the complete specific function

Determine G on the stated domain without choosing from candidate expressions. Derive the logarithmic coefficient, substitute the boundary condition before solving for C, and give both an expanded-log form and an equivalent single-log form. Verify the derivative and the boundary condition independently. Explain why C cannot simply equal −1 and why the original domain must be kept

Specific function task

\(\displaystyle G^{\prime}(x)=9/(3x-2)\)

After deriving your primitive, enter the logarithmic coefficient

Specific function task

\(\displaystyle G^{\prime}(x)=9/(3x-2),\quad G(2)=-1,\quad x=10/3\)

In your single-log form, what is the positive logarithm argument when the input is ten-thirds?

Check both equivalent forms and both conditions

Integration gives G(x)=3 ln(3x−2)+C. At x=2 the logarithm argument is 4, hence C=−1−3 ln 4. Combining logarithms gives G(x)=−1+3 ln((3x−2)/4), on x>2/3. Its derivative is 3×3/(3x−2)=9/(3x−2); at x=2 its value is −1. Differentiation alone cannot detect an incorrect additive constant. The boundary check alone cannot establish the derivative

Transfer the construction to a negative inner branch

Construct H on the stated domain from the derivative and condition below. Begin with an absolute-value logarithm, solve for the nontrivial constant, then give a single logarithm whose argument is positive throughout this branch. Verify the derivative and H(0). Explain why the condition at zero does not fix the constant on the disconnected branch to the right of the pole. Also give the exact function value at negative three-halves

Specific function task

\(\displaystyle H^{\prime}(x)=-10/(2x-3)\)

What is the logarithmic coefficient in your primitive?

Specific function task

\(\displaystyle H^{\prime}(x)=-10/(2x-3),\quad x=0\)

Independently evaluate the given derivative at zero

Preserve the branch when combining logarithms

A primitive is −5 ln|2x−3|+C. At zero, |2x−3|=3, giving C=4+5 ln 3. On x<3/2 the absolute value equals 3−2x, so H(x)=4−5 ln((3−2x)/3). The argument is positive. Its derivative is 10/(3−2x)=−10/(2x−3); at zero the log is zero and H(0)=4. At −3/2 the inner ratio is 2, so H=4−5 ln 2. The pole separates two intervals: a condition on the left does not determine the independent integration constant on the right