What is the integral of ((arctan(x))^4)/(x^2+1) ?

The integral of ((arctan(x))^4)/(x^2+1) is (arctan^5(x))/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((arctan(x))^4)/(x^2+1)

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

\frac{arctan ^{5} ( x )}{5} + C

Solution steps

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

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{u ^{5}}{5}

Substitute back

u = arctan (x)

= \frac{arctan ^{5} ( x )}{5}

Add a constant to the solution

= \frac{arctan ^{5} ( x )}{5} + C

\int x 4 x^{2} + 3 d x

\int 2 t e^{- t} d t

\int 5 sec^{2} (5 x) tan (5 x) d x

\int (3 x - 1) (12 x^{2} - 8 x + 1)^{\frac{1}{3}} d x

\int (2 x^{2} + x^{5}) 2 + x^{3} d x

What is the integral of ((arctan(x))^4)/(x^2+1) ?
The integral of ((arctan(x))^4)/(x^2+1) is (arctan^5(x))/5+C