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

The integral of 1/(sqrt(2x^2+x)) is 1/(sqrt(2))ln|sqrt(16x^2+8x)+4x+1|+C

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

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

\frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1 + C

Solution steps

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

Complete the square

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

= \int \frac{1}{2 ( x + \frac{1}{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{1}{4}))) + sec (arcsec (4 (x + \frac{1}{4})))

Simplify

22 \frac{1}{4} ln tan (arcsec (4 (x + \frac{1}{4}))) + sec (arcsec (4 (x + \frac{1}{4}))) : \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1

= \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1

Add a constant to the solution

= \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1 + C

\frac{d}{d x} ((cos (x))^{s i n (x)})

e^{x}, 14 e^{- x} + 5, x = 0

\int x (4 x^{2} + 8) d x

\int x 1 - x^{4} d x

f (x) = 2 x, at x = 8

What is the integral of 1/(sqrt(2x^2+x)) ?
The integral of 1/(sqrt(2x^2+x)) is 1/(sqrt(2))ln|sqrt(16x^2+8x)+4x+1|+C