integral of (3x^2+1)/(2x) — step-by-step solution

Let's compute the integral:

\int \frac{3 x^{2} + 1}{2 x} d x

For

x

≠ 0, divide each numerator term by the power in the denominator:

\frac{3 x^{2} + 1}{2 x} = \frac{3 x}{2} + \frac{1}{2 x}

ⅰ

\int \frac{3 x^{2} + 1}{2 x} d x

\int \frac{3 x}{2} d x + \int \frac{1}{2 x} d x

\int \frac{3 x}{2} d x + \int \frac{1}{2 x} d x

\int \frac{1}{2 x} d x = \frac{1}{2} \cdot \int \frac{1}{x} d x

We use

\int \frac{1}{x} d x = \ln | x |

. The absolute value covers both signs of the denominator, on any interval where

x \neq 0

ⓘ A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int \frac{3 x}{2} d x

\frac{3}{2} \cdot \frac{x^{2}}{2}

■The derivative of

\frac{1}{2} \cdot \ln | x |

\frac{1}{2 x}

, so

\frac{1}{2} \cdot \ln | x |

is an antiderivative of

\frac{1}{2 x}

\int \frac{1}{2 x} d x

\frac{1}{2} \cdot \ln | x |

\frac{3}{2} \cdot \frac{x^{2}}{2} + \frac{1}{2} \cdot \ln | x |

ⓘ To multiply fractions, multiply the numerators together and the denominators together:

\frac{3}{2} \cdot \frac{x^{2}}{2}

\frac{3 x^{2}}{2 \cdot 2}

\frac{3 x^{2}}{4}

■Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \ln | x |

\frac{1 \cdot \ln | x |}{2}

\frac{\ln | x |}{2}

\frac{3 x^{2}}{4} + \frac{\ln | x |}{2}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\frac{3 x^{2}}{4} + \frac{\ln | x |}{2} + C