What is the integral of 1/(sqrt(2+7y^2)) ?

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

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integral of 1/(sqrt(2+7y^2))

\int \frac{1}{2 + 7 y ^{2}} d y

\frac{1}{7} ln \frac{7}{2} y + \frac{1}{2} (2 + 7 y^{2}) + C

Solution steps

\int \frac{1}{2 + 7 y ^{2}} d y

Apply Trigonometric Substitution

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

= \int \frac{sec ( u )}{7} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{7}{2} y)

= \frac{1}{7} ln tan (arctan (\frac{7}{2} y)) + sec (arctan (\frac{7}{2} y))

Simplify

\frac{1}{7} ln tan (arctan (\frac{7}{2} y)) + sec (arctan (\frac{7}{2} y)) : \frac{1}{7} ln \frac{7}{2} y + \frac{1}{2} (2 + 7 y^{2})

= \frac{1}{7} ln \frac{7}{2} y + \frac{1}{2} (2 + 7 y^{2})

Add a constant to the solution

= \frac{1}{7} ln \frac{7}{2} y + \frac{1}{2} (2 + 7 y^{2}) + C

e^{- 2 t}

\frac{d}{d x} (\frac{5}{x + 4})

\int - 4 e^{- 2 x} d x

\int_{0}^{1} 3 x^{2} + 1 d x

2 + 7 t

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