What is the integral of 7/(sqrt(7x^2+14x+5)) ?

The integral of 7/(sqrt(7x^2+14x+5)) is sqrt(7)ln|sqrt(1/2 (7x^2+14x+5))+sqrt(7/2)(x+1)|+C

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integral of 7/(sqrt(7x^2+14x+5))

\int \frac{7}{7 x ^{2} + 14 x + 5} d x

7 ln \frac{1}{2} (7 x^{2} + 14 x + 5) + \frac{7}{2} (x + 1) + C

Solution steps

\int \frac{7}{7 x ^{2} + 14 x + 5} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{7 x ^{2} + 14 x + 5} d x

Complete the square

7 x^{2} + 14 x + 5 : 7 (x + 1)^{2} - 2

= 7 \cdot \int \frac{1}{7 ( x + 1 ) ^{2} - 2} d x

Apply u-substitution

: \int \frac{1}{7 u ^{2} - 2} d u

= 7 \cdot \int \frac{1}{7 u ^{2} - 2} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{7} d v

= 7 \cdot \int \frac{sec ( v )}{7} d v

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int sec (v) d v

Use the common integral:

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

= 7 \cdot \frac{1}{7} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 7 \cdot \frac{1}{7} ln tan (arcsec (\frac{7}{2} (x + 1))) + sec (arcsec (\frac{7}{2} (x + 1)))

Simplify

7 \cdot \frac{1}{7} ln tan (arcsec (\frac{7}{2} (x + 1))) + sec (arcsec (\frac{7}{2} (x + 1))) : 7 ln \frac{1}{2} (7 x^{2} + 14 x + 5) + \frac{7}{2} (x + 1)

= 7 ln \frac{1}{2} (7 x^{2} + 14 x + 5) + \frac{7}{2} (x + 1)

Add a constant to the solution

= 7 ln \frac{1}{2} (7 x^{2} + 14 x + 5) + \frac{7}{2} (x + 1) + C

x \to 0 lim (\frac{1}{x})

f (x) = \frac{tan ( x ) - 1}{sec ( x )}

\frac{\partial}{\partial y} (e^{y} \cdot cos (x))

\frac{( s + 3 )}{s ^{2} + 2 s + 1}

4 y^{^{''}} - 16 y^{^{'}} + 18 y = 0

What is the integral of 7/(sqrt(7x^2+14x+5)) ?
The integral of 7/(sqrt(7x^2+14x+5)) is sqrt(7)ln|sqrt(1/2 (7x^2+14x+5))+sqrt(7/2)(x+1)|+C