What is the indefinite integral \int \left( (3x + 1)^2 - 6\sin(3x) \right) dx? — JAMB UTME

What is the indefinite integral $\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx$ ?

3x^3 + 3x^2 + x + 2\cos(3x) + C

Expanding

(3x+1)^2

yields

9x^2 + 6x + 1

. Integrating

9x^2 + 6x + 1

gives

3x^3 + 3x^2 + x

. Integrating

-6\sin(3x)

gives

+2\cos(3x)

. Combining these results with the mandatory constant of integration

+ C

gives

3x^3 + 3x^2 + x + 2\cos(3x) + C

Expand the squared polynomial binomial term

(3x + 1)^2 = 9x^2 + 6x + 1

Expanding the expression allows for simple term-by-term integration using the power rule.

Integrate the polynomial component term by term

\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

to each polynomial term.

Integrate the trigonometric component

\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)

The integral of

\sin(kx)

-\frac{1}{k}\cos(kx)

. Multiplying by

-6

yields

+2\cos(3x)

Combine terms and add the constant of integration

3x^3 + 3x^2 + x + 2\cos(3x) + C

Indefinite integrals must include an arbitrary constant

+ C

to account for all antiderivatives.

Indefinite Integration of Polynomial and Trigonometric Functions

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