What is the integral of (a^{arccot(x)}+ax) 1/(1+x^2) ?

The integral of (a^{arccot(x)}+ax) 1/(1+x^2) is aln|sqrt(1+x^2)|-(a^{arccot(x)})/(ln(a))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (a^{arccot(x)}+ax) 1/(1+x^2)

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

a ln 1 + x^{2} - \frac{a ^{arccot (x)}}{ln ( a )} + C

Solution steps

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

Apply u-substitution

: \int - a cot (u) - a^{u} d u

= \int - a cot (u) - a^{u} 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

= - \int a cot (u) d u - \int a^{u} d u

\int a cot (u) d u = a ln ∣ sin (u) ∣

\int a^{u} d u = \frac{a ^{u}}{ln ( a )}

= - a ln ∣ sin (u) ∣ - \frac{a ^{u}}{ln ( a )}

Substitute back

u = arccot (x)

= - a ln ∣ sin (arccot (x)) ∣ - \frac{a ^{arccot (x)}}{ln ( a )}

Simplify

- a ln ∣ sin (arccot (x)) ∣ - \frac{a ^{arccot (x)}}{ln ( a )} : a ln 1 + x^{2} - \frac{a ^{arccot (x)}}{ln ( a )}

= a ln 1 + x^{2} - \frac{a ^{arccot (x)}}{ln ( a )}

Add a constant to the solution

= a ln 1 + x^{2} - \frac{a ^{arccot (x)}}{ln ( a )} + C

\int 10 x^{9} - 12 x^{3} - 5 d x

\int 7 x (3 x^{2} - 2)^{6} d x

\int sec^{13} (2 x) tan^{3} (2 x) d x

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

What is the integral of (a^{arccot(x)}+ax) 1/(1+x^2) ?
The integral of (a^{arccot(x)}+ax) 1/(1+x^2) is aln|sqrt(1+x^2)|-(a^{arccot(x)})/(ln(a))+C