Integrate using partial fractions, identities and parts

Choose a method from the structure

All logarithms here are natural and all angles are in radians. First simplify and look for a known derivative or substitution. For a rational function, divide if needed and use the existing partial-fraction decomposition. For squared circular functions, lower the power with an identity. For a product whose derivative becomes simpler, consider integration by parts. These methods need justification; they are choices, not an automatic rule based on appearance

Separate decomposition from integration

The decomposition lesson finds the coefficients of a rational identity. It does not itself integrate the result. Once decomposed, integrate each term using the appropriate reciprocal, power or inverse-tangent rule. Keep one arbitrary constant on each connected component of the original domain

Work distinct linear factors

Decompose (7x+1)/[(x−1)(x+3)] as 2/(x−1)+5/(x+3): 2(x+3)+5(x−1)=7x+1. A primitive is 2log|x−1|+5log|x+3|+C for x≠1,−3. Absolute values permit either nonzero sign; they do not remove the poles

Give the coefficient of log|x+3| in 2log|x−1|+5log|x+3|+C.

Account for nonmonic factors

For (11x+5)/[(2x−1)(x+2)], the decomposition is 21/[5(2x−1)]+17/[5(x+2)]. The first primitive coefficient is 21/10 because its inner slope is 2. Thus the primitive is (21/10)log|2x−1|+(17/5)log|x+2|+C, with x≠1/2,−2.

A repeated factor changes the integration rule

The decomposition 4x/[(x−1)²(x+1)]=1/(x−1)+2/(x−1)²−1/(x+1) gives log|x−1|−2/(x−1)−log|x+1|+C. The repeated-factor term uses the power rule; only power−1 gives a logarithm. Preserve x≠1,−1.

What coefficient multiplies 1/(x−1) in log|x−1|−2/(x−1)−log|x+1|+C?

Split a quadratic numerator into two useful pieces

For (x²+4x−1)/[(x−1)(x²+1)], the existing decomposition is 2/(x−1)+(−x+3)/(x²+1). Integrating gives 2log|x−1|−(1/2)log(x²+1)+3arctan x+C. The quadratic is positive on R; only x=1 is excluded

Give the coefficient of log(x²+1) in 2log|x−1|−(1/2)log(x²+1)+3arctan x+C.

A shifted quadratic needs its residual constant

For (3x+1)/[(x−2)²+9], write3x+1=(3/2)(2x−4)+7. A primitive is (3/2)log((x−2)²+9)+(7/3)arctan((x−2)/3)+C. Using the original constant1 as the inverse-tangent numerator would miss the shift contribution6.

Keep the polynomial quotient

For [x⁴−3x³+4x²+x+1]/[(x−1)(x²+1)], division and decomposition give x−2+2/(x−1)+(−x+3)/(x²+1). Its primitive is x²/2−2x+2log|x−1|−(1/2)log(x²+1)+3arctan x+C, on each component of x≠1.

Derive the square-reduction identities

From cos2t=1−2sin²t and cos2t=2cos²t−1, solve for the squared functions. These are identities for all real t; no division by sin t or cos t occurs

Which values at x=0 agree with these two graphs and identities?

Integrate after reducing the power

For ∫6sin²(3x) dx, use6sin²(3x)=3−3cos(6x). The primitive is 3x−(1/2)sin(6x)+C. For ∫4cos²(2x) dx, use2+2cos(4x) to get2x+(1/2)sin(4x)+C. Keep the doubled frequency in the reciprocal factor

What coefficient multiplies sin(6x) in 3x−(1/2)sin(6x)+C?

Reduce a product of squares in two stages

Using sin2x=2sin x cos x gives sin²x cos²x=(1/4)sin²2x=(1/8)(1−cos4x). Therefore ∫8sin²x cos²x dx=x−sin4x/4+C. Each identity applies before the integration rule

Derive integration by parts

The product rule is (uv)′=u′v+uv′. Integrate on an interval where these derivatives and integrals are valid, then rearrange. In differential notation, dv=v′dx and du=u′dx

For the integral of x times the exponential of twice x, which choice reduces the remaining integral?

Work one application of parts

For ∫4x e^(−2x) dx, take u=x and dv=4e^(−2x)dx, so du=dx and v=−2e^(−2x). Thus the primitive is −2x e^(−2x)+2∫e^(−2x)dx=−(2x+1)e^(−2x)+C. Differentiate to see the extra exponential terms cancel

For ∫4x e^(−2x) dx, write a primitive as e^(−2x)(Px+Q)+C. Enter Q.

Use the circular sign carefully

For ∫6x sin(3x) dx, choose u=x and dv=6sin(3x)dx, giving v=−2cos(3x). The primitive is −2x cos(3x)+2∫cos(3x)dx=−2x cos(3x)+(2/3)sin(3x)+C.

Repeat parts until the polynomial disappears

For ∫3x²e^x dx, first obtain 3x²e^x−6∫xe^x dx. A second application gives ∫xe^x dx=xe^x−e^x+C. Substituting yields 3e^x(x²−2x+2)+C. Do not lose the sign multiplying the second result

After expanding 3e^x(x²−2x+2)+C, what coefficient multiplies e^x alone?

Differentiate the logarithm in a product

For ∫4x log x dx on x>0, choose u=log x and dv=4x dx. Then du=dx/x and v=2x², so the primitive is 2x²log x−∫2x dx=2x²log x−x²+C. For ∫log x dx, regard the second factor as 1 and choose dv=dx

What coefficient multiplies x² outside the logarithm in 2x²log x−x²+C?

A returning integral can be solved algebraically

Let I=∫e^x sin x dx and J=∫e^x cos x dx. Applying parts gives I=e^x sin x−J and J=e^x cos x+I. Substitute:2I=e^x(sin x−cos x), so I=(1/2)e^x(sin x−cos x)+C. Treat primitive identities up to a constant, combining all arbitrary constants in the final C.

What coefficient multiplies e^x cos x in (1/2)e^x(sin x−cos x)+C?

Parts also has a definite form

For continuously differentiable u and v on [L,R], integrate the product rule across the interval. Both endpoint product terms must be retained

Work the endpoint contribution

For ∫ from 0 to 1 of 2x e^x dx, a primitive is 2e^x(x−1). At 1 it is 0; at 0 it is −2. Therefore the integral is 2. The endpoint x=0 does not make every exponential term vanish

Evaluate ∫ from 0 to 1 of 2x e^x dx exactly

A proper rational definite integral stays on one branch

For ∫ from 2 to 4 of (3x+2)/[x(x+2)] dx, decompose as 1/x+2/(x+2). Both poles lie outside the interval. The result is log2+2log(3/2). A simplified endpoint expression would not justify integrating across a denominator zero

Select and justify a method, ungraded

Find complete primitives for ∫2x/(x²+5) dx, ∫(2x+1)/(x²+x−2) dx, ∫cos²(4x) dx and ∫x cos(2x) dx. Explain why each method is suitable, state the original real domain and differentiate every result. Choosing a method in a quiz does not assess this independent reasoning

Model: four structures give four routes

For ∫2x/(x²+5) dx, recognition or u=x²+5 gives log(x²+5)+C on R. For ∫(2x+1)/(x²+x−2) dx, the numerator is exactly the derivative of the denominator, so recognition gives log|x²+x−2|+C on x≠1,−2; partial fractions 1/(x−1)+1/(x+2) gives the same result. For ∫cos²(4x) dx, square reduction gives x/2+sin(8x)/16+C on R. For ∫x cos(2x) dx, parts gives (x/2)sin2x+(1/4)cos2x+C on R. More than one sound method can exist

Prove both parts formulas, ungraded

Starting from the product rule, derive the indefinite and definite integration-by-parts formulas with their interval assumptions. Explain why integrating a product by multiplying its separate primitives fails

Model: integrate the full product derivative

If u and v are continuously differentiable on an interval, (uv)′=u′v+uv′. An antiderivative of uv′ is uv minus an antiderivative of u′v, up to one constant. On a closed interval [L,R], the fundamental theorem gives the definite formula after subtracting ∫u′v. For a counterexample, multiplying primitives of x and x gives x⁴/4, whose derivative is x³, not x². The correct primitive of x² is x³/3+C.

Explain square reduction and its diagram, ungraded

Derive both squared-function identities from cos2t, use them to integrate sin²(ax) and cos²(ax) for a≠0, and explain the values of sin²x and cos²x at x=0 and their period. Treat a=0 separately rather than dividing by it

Model: retain the frequency condition

The primitives are x/2−sin(2ax)/(4a)+C and x/2+sin(2ax)/(4a)+C. Differentiation returns (1∓cos2ax)/2. For a=0, the original functions are 0 and1, whose primitives are C and x+C. For nonzero a, the squared functions have period π/|a|. At a=1 and x=0, the values are 0 and1. The identity supplies the proof

Handle a repeated quadratic by combining methods, ungraded

Find a primitive of 1/(1+x²)² and verify it. Use the parts identity on ∫1/(1+x²) dx with u=1/(1+x²) and dv=dx. This extension is written practice; the exact generated rational families do not claim every possible rational denominator

Model: solve for the repeated-quadratic integral

Let I=∫1/(1+x²)² dx and J=arctan x. Parts gives J=x/(1+x²)+2∫x²/(1+x²)² dx. Since x²/(1+x²)²=1/(1+x²)−1/(1+x²)², obtain J=x/(1+x²)+2J−2I. Thus I=x/[2(1+x²)]+(1/2)arctan x+C on R. Differentiating the two terms gives (1−x²)/[2(1+x²)²]+1/[2(1+x²)], whose sum is the original integrand

Check a complete solution with technology, ungraded

Use radian-mode technology to differentiate the shifted-quadratic primitive and the cyclic parts primitive above. Numerically integrate 2x e^x on [0,1] and(3x+2)/[x(x+2)] on [2,4]. Compare exact and decimal values, record your own tool operations and explain the original domains

Model: keep verification separate from proof

The first derivatives recover(3x+1)/[(x−2)²+9] and e^x sin x respectively. The definite values are 2 and log2+2log(3/2), approximately 1.50408. The rational interval avoids 0 and−2. A few agreeing decimal values can detect mistakes but cannot prove a symbolic identity, and reading this model does not demonstrate technology execution