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

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

Popular >

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

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

1 + t^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 + t^{2}

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

Simplify

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

= 1 + t^{2}

Add a constant to the solution

= 1 + t^{2} + C

\int (x^{2} - 9 x - 1) d x

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

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

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

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

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