What is the integral of 1/(sqrt(1+25y^2)) ?

The integral of 1/(sqrt(1+25y^2)) is 1/5 ln|5y+sqrt(1+25y^2)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(1+25y^2))

\int \frac{1}{1 + 25 y ^{2}} d y

\frac{1}{5} ln 5 y + 1 + 25 y^{2} + C

Solution steps

\int \frac{1}{1 + 25 y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (5 y)

= \frac{1}{5} ln ∣ tan (arctan (5 y)) + sec (arctan (5 y)) ∣

Simplify

\frac{1}{5} ln ∣ tan (arctan (5 y)) + sec (arctan (5 y)) ∣ : \frac{1}{5} ln 5 y + 1 + 25 y^{2}

= \frac{1}{5} ln 5 y + 1 + 25 y^{2}

Add a constant to the solution

= \frac{1}{5} ln 5 y + 1 + 25 y^{2} + C

\int sec^{5} (\frac{θ}{3}) d θ

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

\int e^{3 x} sin (e^{- x}) d x

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

\int \frac{1}{2 x ^{2} - 7 x + 5} d x

What is the integral of 1/(sqrt(1+25y^2)) ?
The integral of 1/(sqrt(1+25y^2)) is 1/5 ln|5y+sqrt(1+25y^2)|+C