What is the integral of 1/(sqrt(7x)(1+7x)) ?

The integral of 1/(sqrt(7x)(1+7x)) is 2/7 arctan(sqrt(7)sqrt(x))+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(7x)(1+7x))

\int \frac{1}{7 x ( 1 + 7 x )} d x

\frac{2}{7} arctan (7 x) + C

Solution steps

\int \frac{1}{7 x ( 1 + 7 x )} d x

7 x = 7 x

= \int \frac{1}{7 x ( 1 + 7 x )} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{7} \cdot 2 \cdot \frac{1}{7} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{7} \cdot 2 \cdot \frac{1}{7} arctan (v)

Substitute back

= \frac{1}{7} \cdot 2 \cdot \frac{1}{7} arctan (7 x)

Simplify

\frac{1}{7} \cdot 2 \cdot \frac{1}{7} arctan (7 x) : \frac{2}{7} arctan (7 x)

= \frac{2}{7} arctan (7 x)

Add a constant to the solution

= \frac{2}{7} arctan (7 x) + C

\frac{d}{d x} (\frac{1}{g ( x ) ( x ) ^{2} ( x )})

\int \frac{( 4 x - 3 )}{x + 1} d x

x \to 0 lim (- (e^{\frac{1}{x}}))

f (x) = 2 x^{2} + 3

\frac{d}{d x} (x^{(\frac{1}{3})})

What is the integral of 1/(sqrt(7x)(1+7x)) ?
The integral of 1/(sqrt(7x)(1+7x)) is 2/7 arctan(sqrt(7)sqrt(x))+C