What is the integral of 5/(3+7x^2) ?

The integral of 5/(3+7x^2) is 5/(sqrt(21))arctan(sqrt(7/3)x)+C

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integral of 5/(3+7x^2)

\int \frac{5}{3 + 7 x ^{2}} d x

\frac{5}{21} arctan (\frac{7}{3} x) + C

Solution steps

\int \frac{5}{3 + 7 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{3 + 7 x ^{2}} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{21 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{21} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 5 \cdot \frac{1}{21} arctan (u)

Substitute back

u = \frac{7}{3} x

= 5 \cdot \frac{1}{21} arctan (\frac{7}{3} x)

Simplify

5 \cdot \frac{1}{21} arctan (\frac{7}{3} x) : \frac{5}{21} arctan (\frac{7}{3} x)

= \frac{5}{21} arctan (\frac{7}{3} x)

Add a constant to the solution

= \frac{5}{21} arctan (\frac{7}{3} x) + C

\int (cos (x)) (ln (sin (x))) d x

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

x \to 1 lim (x^{2} - 8 x - 1)

f (x) = \frac{e ^{x}}{3 + e ^{x}}

x \to 3 - lim (x^{2} - 9)

What is the integral of 5/(3+7x^2) ?
The integral of 5/(3+7x^2) is 5/(sqrt(21))arctan(sqrt(7/3)x)+C