What is the indefinite integral \int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx? — JAMB UTME

What is the indefinite integral $\int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx$ ?

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Expanding the product

(3 - 2x)(4x + 1)

gives

-8x^2 + 10x + 3

. Integrating term-by-term yields

\int -8x^2 dx = -\frac{8}{3}x^3

\int 10x dx = 5x^2

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = \frac{8}{4}\sin(4x) = 2\sin(4x)

. Adding the integration constant

C

gives the complete result

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Expand the polynomial product inside the integrand

(3 - 2x)(4x + 1) = 12x + 3 - 8x^2 - 2x = -8x^2 + 10x + 3

Expanding the expression allows term-by-term integration using standard rules.

Rewrite the full integrand

\int \left( -8x^2 + 10x + 3 + 8\cos(4x) \right) dx

Substitute the expanded polynomial back into the integral expression.

Integrate each term individually

\int -8x^2 dx = -\frac{8}{3}x^3

\int 10x dx = 5x^2

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

Apply the power rule

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

and linear trigonometric rule

\int \cos(kx) dx = \frac{\sin(kx)}{k}

Combine terms and append the constant of integration

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Indefinite integration requires adding an arbitrary constant

C

Indefinite Integration of Polynomial and Trigonometric Functions

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