What is the integral of ((t^2)sqrt(1+1/(t^2))) ?

The integral of ((t^2)sqrt(1+1/(t^2))) is 1/3 t^2sqrt(t^2+1)+1/3 sqrt(t^2+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((t^2)sqrt(1+1/(t^2)))

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

\frac{1}{3} t^{2} t^{2} + 1 + \frac{1}{3} t^{2} + 1 + C

Solution steps

\int t^{2} 1 + \frac{1}{t ^{2}} d t

Apply Integration By Parts

: \frac{t ^{3} t ^{2} + 1}{3 t ^{2}} - \int - \frac{t ^{2}}{3 t ^{2} + 1} d t

= \frac{t ^{3} t ^{2} + 1}{3 t ^{2}} - \int - \frac{t ^{2}}{3 t ^{2} + 1} d t

\int - \frac{t ^{2}}{3 t ^{2} + 1} d t = - \frac{1}{3} t^{2} + 1

= \frac{t ^{3} t ^{2} + 1}{3 t ^{2}} - (- \frac{1}{3} t^{2} + 1)

Simplify

\frac{t ^{3} t ^{2} + 1}{3 t ^{2}} - (- \frac{1}{3} t^{2} + 1) : \frac{1}{3} t^{2} t^{2} + 1 + \frac{1}{3} t^{2} + 1

= \frac{1}{3} t^{2} t^{2} + 1 + \frac{1}{3} t^{2} + 1

Add a constant to the solution

= \frac{1}{3} t^{2} t^{2} + 1 + \frac{1}{3} t^{2} + 1 + C

\int (\frac{( y ^{3} - 1 )}{y})^{2} d y

\int \frac{8}{( x + 3 ) ^{4}} d x

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

\int cos^{12} (7 t) sin (7 t) d t

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

What is the integral of ((t^2)sqrt(1+1/(t^2))) ?
The integral of ((t^2)sqrt(1+1/(t^2))) is 1/3 t^2sqrt(t^2+1)+1/3 sqrt(t^2+1)+C