What is the integral of 1/(16x^2+7) ?

The integral of 1/(16x^2+7) is 1/(4sqrt(7))arctan(4/(sqrt(7))x)+C

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integral of 1/(16x^2+7)

\int \frac{1}{16 x ^{2} + 7} d x

\frac{1}{4 7} arctan (\frac{4}{7} x) + C

Solution steps

\int \frac{1}{16 x ^{2} + 7} d x

Apply Integral Substitution

= \int \frac{1}{4 7 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{4 7} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{4 7} arctan (u)

Substitute back

u = \frac{4}{7} x

= \frac{1}{4 7} arctan (\frac{4}{7} x)

Add a constant to the solution

= \frac{1}{4 7} arctan (\frac{4}{7} x) + C

\frac{13}{x}

\frac{d}{d t} (e^{7 t s i n (2 t)})

x \to 0 lim (sin (\frac{1}{x}))

\int x csc (x^{2}) cot (x^{2}) d x

\int \frac{1}{( x ^{2} + 25 )} d x

What is the integral of 1/(16x^2+7) ?
The integral of 1/(16x^2+7) is 1/(4sqrt(7))arctan(4/(sqrt(7))x)+C