What is the integral of 1/(-x^2+4x-5) ?

The integral of 1/(-x^2+4x-5) is -arctan(x-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/(-x^2+4x-5)

\int \frac{1}{- x ^{2} + 4 x - 5} d x

- arctan (x - 2) + C

Solution steps

\int \frac{1}{- x ^{2} + 4 x - 5} d x

Complete the square

- x^{2} + 4 x - 5 : - (x - 2)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{- u ^{2} - 1} d u

= \int \frac{1}{- u ^{2} - 1} d u

Factor

- u^{2} - 1 : - (u^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (u)

Substitute back

u = x - 2

= - arctan (x - 2)

Add a constant to the solution

= - arctan (x - 2) + C

\int x cot^{2} (x) d x

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

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

\int e^{x} (2^{x} - 1) d x

What is the integral of 1/(-x^2+4x-5) ?
The integral of 1/(-x^2+4x-5) is -arctan(x-2)+C