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

The integral of 2tsqrt(1+t^2) is 2/3 (1+t^2)^{3/2}+C

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integral of 2tsqrt(1+t^2)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int t 1 + t^{2} d t

Apply u-substitution

: \int u^{2} d u

= 2 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 + t^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{( x ^{2} + 4 x + 13 ) ^{\frac{3}{2}}} d x

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

\frac{d}{d x} (\frac{x}{2 - x ^{2}})

f (x) = \frac{x}{x - 2}, at x = 3

(- 5.2) (2.5)

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