What is the integral of 7/(16+9x^2) ?

The integral of 7/(16+9x^2) is 7/12 arctan(3/4 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7/(16+9x^2)

\int \frac{7}{16 + 9 x ^{2}} d x

\frac{7}{12} arctan (\frac{3}{4} x) + C

Solution steps

\int \frac{7}{16 + 9 x ^{2}} 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}{16 + 9 x ^{2}} d x

Apply Integral Substitution

= 7 \cdot \int \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 7 \cdot \frac{1}{12} arctan (u)

Substitute back

u = \frac{3}{4} x

= 7 \cdot \frac{1}{12} arctan (\frac{3}{4} x)

Simplify

7 \cdot \frac{1}{12} arctan (\frac{3}{4} x) : \frac{7}{12} arctan (\frac{3}{4} x)

= \frac{7}{12} arctan (\frac{3}{4} x)

Add a constant to the solution

= \frac{7}{12} arctan (\frac{3}{4} x) + C

\int x \cdot cos (π x) d x

\int (- 7 x + 5)^{3} d x

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

\int (1 + t) e^{t} d t

\int cot (x) csc^{4} (x) d x

What is the integral of 7/(16+9x^2) ?
The integral of 7/(16+9x^2) is 7/12 arctan(3/4 x)+C