What is the integral of (t^6)/(sqrt(1-t^{14))} ?

The integral of (t^6)/(sqrt(1-t^{14))} is 1/7 arcsin(t^7)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (t^6)/(sqrt(1-t^{14))}

\int \frac{t ^{6}}{1 - t ^{14}} d t

\frac{1}{7} arcsin (t^{7}) + C

Solution steps

\int \frac{t ^{6}}{1 - t ^{14}} d t

Apply u-substitution

= \int \frac{1}{7 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}{7} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= \frac{1}{7} arcsin (u)

Substitute back

u = t^{7}

= \frac{1}{7} arcsin (t^{7})

Add a constant to the solution

= \frac{1}{7} arcsin (t^{7}) + C

What is the integral of (t^6)/(sqrt(1-t^{14))} ?
The integral of (t^6)/(sqrt(1-t^{14))} is 1/7 arcsin(t^7)+C