What is the integral of 31arctan(sqrt(x)) ?

The integral of 31arctan(sqrt(x)) is 62(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 31arctan(sqrt(x))

\int 31 arctan (x) d x

62 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

Solution steps

\int 31 arctan (x) d x

Take the constant out:

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

= 31 \cdot \int arctan (x) d x

Apply u-substitution

: \int arctan (u) \cdot 2 u d u

= 31 \cdot \int arctan (u) \cdot 2 u d u

Take the constant out:

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

= 31 \cdot 2 \cdot \int arctan (u) u d u

Apply Integration By Parts

: \frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u

= 31 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)

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

= 31 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 31 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

31 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 62 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 62 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 62 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

\int_{1}^{4} 3 x^{\frac{1}{3}} d x

\frac{d}{d x} (lo g_{2} (7 x^{2} + x - 2))

\frac{d}{d x} (12 x - x^{2})

x \to 1 lim (\frac{( x + 1 )}{x - 1})

\frac{d}{d x} (x^{2} - \frac{1}{8} ln (x))

What is the integral of 31arctan(sqrt(x)) ?
The integral of 31arctan(sqrt(x)) is 62(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C