What is the integral of (x^2)/(1+16x^2) ?

The integral of (x^2)/(1+16x^2) is 1/64 (-arctan(4x)+4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(1+16x^2)

\int \frac{x ^{2}}{1 + 16 x ^{2}} d x

\frac{1}{64} (- arctan (4 x) + 4 x) + C

Solution steps

\int \frac{x ^{2}}{1 + 16 x ^{2}} d x

Apply Integral Substitution

= \int \frac{u ^{2}}{64 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

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

= \frac{1}{64} \cdot \int - \frac{1}{1 + u ^{2}} + 1 d u

Apply the Sum Rule:

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

= \frac{1}{64} (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

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

\int 1 d u = u

= \frac{1}{64} (- arctan (u) + u)

Substitute back

u = 4 x

= \frac{1}{64} (- arctan (4 x) + 4 x)

Add a constant to the solution

= \frac{1}{64} (- arctan (4 x) + 4 x) + C

\int 10 sin^{2} (\frac{5 x}{2}) d x

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

\int 5^{2 x^{3}} 2 x^{2} d x

\int - 4 sin (2 z) d z

\int 6 x e^{6 x} d x

What is the integral of (x^2)/(1+16x^2) ?
The integral of (x^2)/(1+16x^2) is 1/64 (-arctan(4x)+4x)+C