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

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

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

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

- ln ∣ 4 x - 4 ∣ + ln ∣ 4 x - 6 ∣ + 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

= \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 - 4 ∣ + ln ∣ 4 x - 6 ∣

= - ln ∣ 4 x - 4 ∣ + ln ∣ 4 x - 6 ∣

Add a constant to the solution

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

\int_{0}^{1} x^{\frac{1}{3}} - x^{3} d x

\int x 9 36 + x^{2} d x

\int cos (3 x) cos (2 x) cos (x) d x

y^{^{'}} = \frac{( 6 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2} )}{x ^{3} y}

n = 0 \sum \infty \frac{1}{( n ^{2} - 1 )}

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