What is the integral of 1/((2x-3)^2+1) ?

The integral of 1/((2x-3)^2+1) is 1/2 arctan(2x-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((2x-3)^2+1)

\int \frac{1}{( 2 x - 3 ) ^{2} + 1} d x

\frac{1}{2} arctan (2 x - 3) + C

Solution steps

\int \frac{1}{( 2 x - 3 ) ^{2} + 1} d x

Expand

(2 x - 3)^{2} + 1 : 4 x^{2} - 12 x + 10

= \int \frac{1}{4 x ^{2} - 12 x + 10} d x

Complete the square

4 x^{2} - 12 x + 10 : 4 (x - \frac{3}{2})^{2} + 1

= \int \frac{1}{4 ( x - \frac{3}{2} ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 1} d u

= \int \frac{1}{4 u ^{2} + 1} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{2} arctan (2 (x - \frac{3}{2}))

Simplify

\frac{1}{2} arctan (2 (x - \frac{3}{2})) : \frac{1}{2} arctan (2 x - 3)

= \frac{1}{2} arctan (2 x - 3)

Add a constant to the solution

= \frac{1}{2} arctan (2 x - 3) + C

x \to 5 lim (5)

\int \frac{( 3 + x ) ^{2}}{x} d x

\int 3 x 2 x + 3 d x

\int cos^{3} (\frac{x}{8}) d x

y = \frac{1}{x ^{5}}, (3, \frac{1}{243})

What is the integral of 1/((2x-3)^2+1) ?
The integral of 1/((2x-3)^2+1) is 1/2 arctan(2x-3)+C