What is the integral of 1/((4x+1)sqrt(16x^2+4)) ?

The integral of 1/((4x+1)sqrt(16x^2+4)) is

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integral of 1/((4x+1)sqrt(16x^2+4))

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

\frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1 + C

Solution steps

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{4 u u ^{2} - 2 u + 5} d u

= \int \frac{1}{4 u u ^{2} - 2 u + 5} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u u ^{2} - 2 u + 5} d u

Complete the square

u^{2} - 2 u + 5 : (u - 1)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{( v + 1 ) v ^{2} + 4} d v

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

Apply Trigonometric Substitution

: \int \frac{sec ( w )}{2 tan ( w ) + 1} d w

= \frac{1}{4} \cdot \int \frac{sec ( w )}{2 tan ( w ) + 1} d w

Apply u-substitution

: \int \frac{2}{4 t + 1 - t ^{2}} d t

= \frac{1}{4} \cdot \int \frac{2}{4 t + 1 - t ^{2}} d t

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{4 t + 1 - t ^{2}} d t

Factor

4 t + 1 - t^{2} : - (- 4 t - 1 + t^{2})

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

Take the constant out:

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

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

Complete the square

- 4 t - 1 + t^{2} : (t - 2)^{2} - 5

= \frac{1}{4} \cdot 2 (- \int \frac{1}{( t - 2 ) ^{2} - 5} d t)

Apply u-substitution

: \int \frac{1}{r ^{2} - 5} d r

= \frac{1}{4} \cdot 2 (- \int \frac{1}{r ^{2} - 5} d r)

Apply Integral Substitution

= \frac{1}{4} \cdot 2 (- \int \frac{1}{5 ( s ^{2} - 1 )} d s)

Take the constant out:

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} \cdot \int \frac{1}{s ^{2} - 1} d s)

Factor

s^{2} - 1 : - (- s^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} (- (\frac{ln ∣ s + 1 ∣}{2} - \frac{ln ∣ s - 1 ∣}{2})))

Substitute back

= \frac{1}{4} \cdot 2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} - 1}{2}

Simplify

\frac{1}{4} \cdot 2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} - 1}{2} : \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1

= \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1

Add a constant to the solution

= \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1 + C

f (x) = - 1 - 8 x^{2}, at x = - 2

\int_{0}^{1} (x) - (x^{2}) d x

\int (x - 7)^{2} d x

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

x \to \infty lim (0^{0})

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