What is the integral of (48x)/(16x^2+8x+10) ?

The integral of (48x)/(16x^2+8x+10) is 3/2 ln|16x^2+8x+10|-arctan(4/3 x+1/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (48x)/(16x^2+8x+10)

\int \frac{48 x}{16 x ^{2} + 8 x + 10} d x

\frac{3}{2} ln 16 x^{2} + 8 x + 10 - arctan (\frac{4}{3} x + \frac{1}{3}) + C

Solution steps

\int \frac{48 x}{16 x ^{2} + 8 x + 10} d x

Take the constant out:

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

= 48 \cdot \int \frac{x}{16 x ^{2} + 8 x + 10} d x

Complete the square

16 x^{2} + 8 x + 10 : 16 (x + \frac{1}{4})^{2} + 9

= 48 \cdot \int \frac{x}{16 ( x + \frac{1}{4} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{4 u - 1}{4 ( 16 u ^{2} + 9 )} d u

= 48 \cdot \int \frac{4 u - 1}{4 ( 16 u ^{2} + 9 )} d u

Take the constant out:

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

= 48 \cdot \frac{1}{4} \cdot \int \frac{4 u - 1}{16 u ^{2} + 9} d u

Expand

\frac{4 u - 1}{16 u ^{2} + 9} : \frac{4 u}{16 u ^{2} + 9} - \frac{1}{16 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

= 48 \cdot \frac{1}{4} (\int \frac{4 u}{16 u ^{2} + 9} d u - \int \frac{1}{16 u ^{2} + 9} d u)

\int \frac{4 u}{16 u ^{2} + 9} d u = \frac{1}{8} ln 16 u^{2} + 9

\int \frac{1}{16 u ^{2} + 9} d u = \frac{1}{12} arctan (\frac{4}{3} u)

= 48 \cdot \frac{1}{4} (\frac{1}{8} ln 16 u^{2} + 9 - \frac{1}{12} arctan (\frac{4}{3} u))

Substitute back

u = x + \frac{1}{4}

= 48 \cdot \frac{1}{4} (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 9 - \frac{1}{12} arctan (\frac{4}{3} (x + \frac{1}{4})))

Simplify

48 \cdot \frac{1}{4} (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 9 - \frac{1}{12} arctan (\frac{4}{3} (x + \frac{1}{4}))) : \frac{3}{2} ln 16 x^{2} + 8 x + 10 - arctan (\frac{4}{3} x + \frac{1}{3})

= \frac{3}{2} ln 16 x^{2} + 8 x + 10 - arctan (\frac{4}{3} x + \frac{1}{3})

Add a constant to the solution

= \frac{3}{2} ln 16 x^{2} + 8 x + 10 - arctan (\frac{4}{3} x + \frac{1}{3}) + C

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

\int sin (3 y) cos (5 y) d y

\int 1682 x^{2} + 987 x + 4774 d x

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

What is the integral of (48x)/(16x^2+8x+10) ?
The integral of (48x)/(16x^2+8x+10) is 3/2 ln|16x^2+8x+10|-arctan(4/3 x+1/3)+C