Integration by recognition with logarithmic components

What should be recognised before integrating the target?

Inner function

\(\displaystyle g(x)=x^2+4\)

Change k and compare the unchanged inner function with the scaled integrand. Table values may be displayed as decimals

What does changing k affect?

Integrate by logarithmic recognition

When the numerator is the derivative of an inner function, the quotient has a logarithmic anti-derivative. A constant multiplying the quotient remains as the logarithmic coefficient

1. 1Find the inner function Identify the denominator or other whole function whose derivative appears in the numerator \(\displaystyle g(x)=x^{2}+4,\quad g'(x)=2x\)
2. 2Match the whole quotient The numerator must be the derivative of the complete inner function, not just part of it \(\displaystyle \int\frac{2x}{x^{2}+4}\,dx=\log_e\left|x^{2}+4\right|+C\)
3. 3Carry the constant scale A constant outside the derivative-over-function quotient stays outside the logarithm \(\displaystyle \int5\frac{g'(x)}{g(x)}\,dx=5\log_e\left|g(x)\right|+C\)
4. 4Differentiate to check Differentiate the proposed logarithm and compare it with the original quotient \(\displaystyle \frac{d}{dx}\left(3\log_e\left|g(x)\right|\right)=3\frac{g'(x)}{g(x)}\)

Function data

\(\displaystyle g(x)=x^2-3x+7\)

Differentiate the inner function and find the coefficient of x in its derivative

Function data

\(\displaystyle f(x)=4\frac{2x-3}{x^2-3x+7}\)

Read the scale multiplying the recognised derivative

Reporting the derivative. One of these lines is right. Which one?

Match the whole inner derivative

Differentiate the complete denominator before finding the constant multiplier. The positive completed-square form also establishes the real domain

1. 1Keep the original inner function The inner function is positive on the whole real line \(\displaystyle g(x)=2x^2-4x+5=2(x-1)^2+3\)
2. 2Compare the numerator The polynomial identity holds even at x=1, where both sides vanish; no division by zero is needed \(\displaystyle g^{\prime}(x)=4x-4,\qquad 10x-10=\frac52(4x-4)\)
3. 3Integrate and check Differentiation recovers the whole integrand. Keep the supplied inner function unchanged when entering K. \(\displaystyle \int\frac{10x-10}{2x^2-4x+5}\,dx=\frac52\ln(2x^2-4x+5)+C\)

Function data

\(\displaystyle \int\frac{10x-10}{2x^2-4x+5}\,dx=K\ln(2x^2-4x+5)+C\)

Find K in the stated primitive form. Keep the inner quadratic unchanged

Recognise a supplied product derivative

Use both terms of the product rule. Matching a complete derivative is not a rule for integrating factors separately

1. 1Define the supplied function The quadratic is positive for every real x. \(\displaystyle H(x)=(2x-1)\ln(x^2+4)\)
2. 2Differentiate both factors The first term comes from differentiating 2x−1; the second comes from differentiating the logarithm \(\displaystyle H^{\prime}(x)=2\ln(x^2+4)+(2x-1)\frac{2x}{x^2+4}\)
3. 3Match the target Every term of this target is three times the corresponding term of H′ \(\displaystyle f(x)=6\ln(x^2+4)+\frac{6x(2x-1)}{x^2+4}=3H^{\prime}(x)\)
4. 4Read off the whole primitive The arbitrary constant supplies the complete family on the real line \(\displaystyle \int f(x)\,dx=3(2x-1)\ln(x^2+4)+C\)

Function data

\(\displaystyle f(x)=6\ln(x^2+4)+\frac{6x(2x-1)}{x^2+4}\)

Primitive form

\(\displaystyle \int f(x)\,dx=K(2x-1)\ln(x^2+4)+C\)

Find K in the stated primitive form. Keep both supplied factors unchanged

Why must the derivative of (2x−1) ln(x²+4) include 2 ln(x²+4)?

Keep the original endpoint ratio

Use the primitive throughout the integration interval. For a single logarithmic term, the entry convention keeps K unchanged and uses the upper inner value divided by the lower inner value

1. 1Find a primitive The inner derivative is 2x, so the coefficient is 3. \(\displaystyle F(x)=3\ln(x^2+4),\qquad F^{\prime}(x)=\frac{6x}{x^2+4}\)
2. 2Evaluate in order The original denominator stays positive on [0,2] \(\displaystyle \int_0^2\frac{6x}{x^2+4}\,dx=3\ln8-3\ln4\)
3. 3Combine with the coefficient fixed The specified entries are K=3 and R=2. Moving the factor 3 into the logarithm gives an equivalent value but not the requested entry form \(\displaystyle 3\ln8-3\ln4=3\ln(8/4)=3\ln2\)

Function data

\(\displaystyle \int_0^2\frac{6x}{x^2+4}\,dx=K\ln(R)\)

Find R in K ln(R), keeping K equal to the primitive coefficient and using the original denominator at the endpoints

Keep two signed endpoint terms

Use the stated primitive for this target. The first logarithmic term is the upper primitive value; the second is the negative of the lower primitive value. Their coefficients need not be equal

1. 1Upper endpoint Substitute into both the linear factor and the inner quadratic \(\displaystyle F(2)=3(3)\ln8=9\ln8\)
2. 2Lower endpoint The negative linear factor makes this primitive value negative \(\displaystyle F(0)=3(-1)\ln4=-3\ln4\)
3. 3Subtract the lower value The second coefficient is positive because the lower primitive value is subtracted. Keep the arguments 8 and 4 in endpoint order \(\displaystyle \int_0^2 f(x)\,dx=F(2)-F(0)=9\ln8+3\ln4\)

Function data

\(\displaystyle F(x)=3(2x-1)\ln(x^2+4)\)

Enter the coefficient of the first logarithmic term in F(2)−F(0), keeping g(2) as its argument

Function data

\(\displaystyle F(x)=3(2x-1)\ln(x^2+4)\)

Enter the signed coefficient of the second logarithmic term in F(2)−F(0), keeping g(0) as its argument

Construct a complete primitive, ungraded

Without selecting from candidate expressions, find the full primitive family on the real line. Identify the inner derivative, derive the scale, prove the logarithm is defined and differentiate your final expression. Write the whole function on paper

Check the construction and domain

Compare your complete function with the model, including the arbitrary constant and the derivative verification

1. 1Inner function and scale The numerator is five times the entire derivative \(\displaystyle g(x)=x^2+3x+8,\quad g^{\prime}(x)=2x+3,\quad 10x+15=5(2x+3)\)
2. 2Complete family and domain The completed square has positive minimum 23/4, so the formula holds for every real x. \(\displaystyle F(x)=5\ln(x^2+3x+8)+C,\quad g(x)=(x+3/2)^2+23/4\)
3. 3Differentiate The derivative matches the original target, including its scale \(\displaystyle F^{\prime}(x)=5\frac{2x+3}{x^2+3x+8}=\frac{10x+15}{x^2+3x+8}\)

Transfer the whole-derivative method, ungraded

Use the supplied product to construct the complete primitive family for h. Differentiate J, compare every term, state the domain and evaluate the definite integral from 0 to 1 in ordered logarithmic terms. Record the whole derivation on paper

Function data

\(\displaystyle h(x)=-2\ln(x^2+3)-\frac{4x(x+2)}{x^2+3}\)

After deriving your primitive, enter K in K(x+2) ln(x²+3)+C. Keep both factors unchanged

Check both terms and both endpoints

The inner quadratic is positive on the real line, including the entire interval [0,1]. An endpoint calculation alone would not establish the derivative identity

1. 1Differentiate the supplied product Keep both product-rule terms \(\displaystyle J^{\prime}(x)=\ln(x^2+3)+(x+2)\frac{2x}{x^2+3}\)
2. 2Construct and verify Since h=−2J′, differentiation of this family recovers h. \(\displaystyle \int h(x)\,dx=-2(x+2)\ln(x^2+3)+C\)
3. 3Use upper minus lower The ordered coefficients are −6 and +4; the arguments are 4 and 3. \(\displaystyle \int_0^1h(x)\,dx=-6\ln4-(-4\ln3)=-6\ln4+4\ln3\)

Check the original interval before subtracting endpoints

These examples have positive quadratic inner functions, so their logarithms and primitives are valid on the whole real line. In a more general derivative-over-function problem, use the displayed logarithmic rule on each connected interval where the inner function is non-zero. Endpoint subtraction across a zero of g is not justified by this rule. Constants may differ between disconnected intervals. Complete written functions, domain explanations and independent derivations remain ungraded