What is the solution for f^'(x)=-(4x)/((1+2x^2)^2) ?

The solution for f^'(x)=-(4x)/((1+2x^2)^2) is f(x)= 1/(1+2x^2)+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f^'(x)=-(4x)/((1+2x^2)^2)

f^{^{'}} (x) = - \frac{4 x}{( 1 + 2 x ^{2} ) ^{2}}

f (x) = \frac{1}{1 + 2 x ^{2}} + c_{1}

Solution steps

f^{'} (x) = - \frac{4 x}{( 1 + 2 x ^{2} ) ^{2}}

f^{^{'}} (x) = g (x)

then

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

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

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

f (x) = \frac{1}{1 + 2 x ^{2}} + c_{1}

(6 y + 2 x - 7) d x + (4 y + 6 x - 1) d y = 0

\int_{- 1}^{1} 2 - ∣ x ∣ d x

\int \frac{1}{e ^{- t}} d t

\int 4 cos^{3} (6 x) d x

\int x \cdot sin (k \cdot x) d x

What is the solution for f^'(x)=-(4x)/((1+2x^2)^2) ?
The solution for f^'(x)=-(4x)/((1+2x^2)^2) is f(x)= 1/(1+2x^2)+c_{1}