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

The integral of 1/(sqrt(2x^2+5x+3)) is 1/(sqrt(2))ln|sqrt(16x^2+40x+24)+4x+5|+C

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

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

\frac{1}{2} ln 16 x^{2} + 40 x + 24 + 4 x + 5 + 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{2 2}{16 u ^{2} - 1} d u

= \int \frac{2 2}{16 u ^{2} - 1} 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} - 1} 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 (4 (x + \frac{5}{4}))) + sec (arcsec (4 (x + \frac{5}{4})))

Simplify

22 \frac{1}{4} ln tan (arcsec (4 (x + \frac{5}{4}))) + sec (arcsec (4 (x + \frac{5}{4}))) : \frac{1}{2} ln 16 x^{2} + 40 x + 24 + 4 x + 5

= \frac{1}{2} ln 16 x^{2} + 40 x + 24 + 4 x + 5

Add a constant to the solution

= \frac{1}{2} ln 16 x^{2} + 40 x + 24 + 4 x + 5 + C

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What is the integral of 1/(sqrt(2x^2+5x+3)) ?
The integral of 1/(sqrt(2x^2+5x+3)) is 1/(sqrt(2))ln|sqrt(16x^2+40x+24)+4x+5|+C