What is the integral of (x+1)/(x^2+7x+10) ?

The integral of (x+1)/(x^2+7x+10) is 2(1/4 ln|4x^2+28x+40|+5/12 (ln| 10/3+2/3 x|-ln| 4/3+2/3 x|))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+1)/(x^2+7x+10)

\int \frac{x + 1}{x ^{2} + 7 x + 10} d x

2 (\frac{1}{4} ln 4 x^{2} + 28 x + 40 + \frac{5}{12} (ln \frac{10}{3} + \frac{2}{3} x - ln \frac{4}{3} + \frac{2}{3} x)) + C

Solution steps

\int \frac{x + 1}{x ^{2} + 7 x + 10} d x

Complete the square

x^{2} + 7 x + 10 : (x + \frac{7}{2})^{2} - \frac{9}{4}

= \int \frac{x + 1}{( x + \frac{7}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 5 )}{4 u ^{2} - 9} d u

= \int \frac{2 ( 2 u - 5 )}{4 u ^{2} - 9} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u - 5}{4 u ^{2} - 9} d u

Expand

\frac{2 u - 5}{4 u ^{2} - 9} : \frac{2 u}{4 u ^{2} - 9} - \frac{5}{4 u ^{2} - 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{2 u}{4 u ^{2} - 9} d u - \int \frac{5}{4 u ^{2} - 9} d u)

\int \frac{2 u}{4 u ^{2} - 9} d u = \frac{1}{4} ln 4 u^{2} - 9

\int \frac{5}{4 u ^{2} - 9} d u = - \frac{5}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 9 - (- \frac{5}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1)))

Substitute back

u = x + \frac{7}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 9 - (- \frac{5}{12} (ln \frac{2}{3} (x + \frac{7}{2}) + 1 - ln \frac{2}{3} (x + \frac{7}{2}) - 1)))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 9 - (- \frac{5}{12} (ln \frac{2}{3} (x + \frac{7}{2}) + 1 - ln \frac{2}{3} (x + \frac{7}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 40 + \frac{5}{12} (ln \frac{10}{3} + \frac{2}{3} x - ln \frac{4}{3} + \frac{2}{3} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 40 + \frac{5}{12} (ln \frac{10}{3} + \frac{2}{3} x - ln \frac{4}{3} + \frac{2}{3} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 40 + \frac{5}{12} (ln \frac{10}{3} + \frac{2}{3} x - ln \frac{4}{3} + \frac{2}{3} x)) + C

\int 12 - 8 x + x^{2} d x

\frac{d ^{2} S}{d x ^{2}} = S

x \to - \infty lim (- 3 x^{3} - x)

(2 x - 6)^{4} (x^{2} + x + 1)^{5}

\frac{d}{d t} (t - t^{- 1})

What is the integral of (x+1)/(x^2+7x+10) ?
The integral of (x+1)/(x^2+7x+10) is 2(1/4 ln|4x^2+28x+40|+5/12 (ln| 10/3+2/3 x|-ln| 4/3+2/3 x|))+C