What is the integral of (35x^2)/((x+1)(x^2+1)) ?

The integral of (35x^2)/((x+1)(x^2+1)) is 35(1/2 ln|x+1|+1/2 (1/2 ln|x^2+1|-arctan(x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (35x^2)/((x+1)(x^2+1))

\int \frac{35 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

35 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) + C

Solution steps

\int \frac{35 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the constant out:

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

= 35 \cdot \int \frac{x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} : \frac{1}{2 ( x + 1 )} + \frac{x - 1}{2 ( x ^{2} + 1 )}

= 35 \cdot \int \frac{1}{2 ( x + 1 )} + \frac{x - 1}{2 ( x ^{2} + 1 )} d x

Apply the Sum Rule:

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

= 35 (\int \frac{1}{2 ( x + 1 )} d x + \int \frac{x - 1}{2 ( x ^{2} + 1 )} d x)

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

= 35 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

Add a constant to the solution

= 35 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) + C

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

\int x 1 5^{x} d x

\int \frac{4 x ^{2} + 29}{x ^{3} + 6 x ^{2} + 9 x - 50} d x

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

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

What is the integral of (35x^2)/((x+1)(x^2+1)) ?
The integral of (35x^2)/((x+1)(x^2+1)) is 35(1/2 ln|x+1|+1/2 (1/2 ln|x^2+1|-arctan(x)))+C