What is the integral of 1/(2x^2+5x+3) ?

The integral of 1/(2x^2+5x+3) is -ln|4x+6|+ln|4x+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(2x^2+5x+3)

\int \frac{1}{2 x ^{2} + 5 x + 3} d x

- ln ∣ 4 x + 6 ∣ + ln ∣ 4 x + 4 ∣ + C

Solution steps

\int \frac{1}{2 x ^{2} + 5 x + 3} d x

Complete the square

2 x^{2} + 5 x + 3 : 2 (x + \frac{5}{4})^{2} - \frac{1}{8}

= \int \frac{1}{2 ( x + \frac{5}{4} ) ^{2} - \frac{1}{8}} d x

Apply u-substitution

= \int \frac{8}{16 u ^{2} - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{16 u ^{2} - 1} d u

Factor

16 u^{2} - 1 : - (- 16 u^{2} + 1)

= 8 \cdot \int \frac{1}{- ( - 16 u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- \int \frac{1}{- 16 u ^{2} + 1} d u)

Apply Integral Substitution

= 8 (- \int \frac{1}{4 ( - v ^{2} + 1 )} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- \frac{1}{4} \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 (- \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{5}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{5}{4} ) - 1}{2}))

Simplify

8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{5}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{5}{4} ) - 1}{2})) : - ln ∣ 4 x + 6 ∣ + ln ∣ 4 x + 4 ∣

= - ln ∣ 4 x + 6 ∣ + ln ∣ 4 x + 4 ∣

Add a constant to the solution

= - ln ∣ 4 x + 6 ∣ + ln ∣ 4 x + 4 ∣ + C

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

\int 84 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int (x - x^{2}) e^{3 x} d x

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

\int \frac{e ^{5 x}}{9 + e ^{5 x}} d x

What is the integral of 1/(2x^2+5x+3) ?
The integral of 1/(2x^2+5x+3) is -ln|4x+6|+ln|4x+4|+C