What is the integral of 1/(sqrt(2x^2-7x+5)) ?

The integral of 1/(sqrt(2x^2-7x+5)) is 1/(sqrt(2))ln((|sqrt(16x^2-56x+40)+4x-7|)/3)+C

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integral of 1/(sqrt(2x^2-7x+5))

\int \frac{1}{2 x ^{2} - 7 x + 5} d x

\frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3}) + C

Solution steps

\int \frac{1}{2 x ^{2} - 7 x + 5} d x

Complete the square

2 x^{2} - 7 x + 5 : 2 (x - \frac{7}{4})^{2} - \frac{9}{8}

= \int \frac{1}{2 ( x - \frac{7}{4} ) ^{2} - \frac{9}{8}} d x

Apply u-substitution

: \int \frac{2 2}{16 u ^{2} - 9} d u

= \int \frac{2 2}{16 u ^{2} - 9} d u

Take the constant out:

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

= 22 \cdot \int \frac{1}{16 u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{4} d v

= 22 \cdot \int \frac{sec ( v )}{4} d v

Take the constant out:

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

= 22 \frac{1}{4} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 22 \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 22 \frac{1}{4} ln tan (arcsec (\frac{4}{3} (x - \frac{7}{4}))) + sec (arcsec (\frac{4}{3} (x - \frac{7}{4})))

Simplify

22 \frac{1}{4} ln tan (arcsec (\frac{4}{3} (x - \frac{7}{4}))) + sec (arcsec (\frac{4}{3} (x - \frac{7}{4}))) : \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3})

= \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3})

Add a constant to the solution

= \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3}) + C

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What is the integral of 1/(sqrt(2x^2-7x+5)) ?
The integral of 1/(sqrt(2x^2-7x+5)) is 1/(sqrt(2))ln((|sqrt(16x^2-56x+40)+4x-7|)/3)+C