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Popular >

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

Solution

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

Solution

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = 4 x

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

Add a constant to the solution

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

Popular Examples

integral of (5x-3)

\int (5 x - 3) d x

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\int \frac{2 x ^{2} - 8 x - 8}{( x - 2 ) ( x ^{2} + 4 )} d x

integral of (7^{2/3}-x^{2/3})^3

\int (7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} d x

integral of 3cos^5(2x)

\int 3 cos^{5} (2 x) d x

integral of (ln(x^3))/(x^2)

\int \frac{ln ( x ^{3} )}{x ^{2}} d x

Frequently Asked Questions (FAQ)

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