What is the integral from 0 to 1 of 1/((mx+1)^2) ?

The integral from 0 to 1 of 1/((mx+1)^2) is 1/(1+m)

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integral from 0 to 1 of 1/((mx+1)^2)

\int_{0}^{1} \frac{1}{( m x + 1 ) ^{2}} d x

\frac{1}{1 + m}

Solution steps

\int_{0}^{1} \frac{1}{( m x + 1 ) ^{2}} d x

Apply u-substitution

: \int_{1}^{1 + m} \frac{1}{m u ^{2}} d u

= \int_{1}^{1 + m} \frac{1}{m u ^{2}} d u

Take the constant out:

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

= \frac{1}{m} \cdot \int_{1}^{1 + m} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{1 + m}

= \frac{1}{m} [- \frac{1}{u}]_{1}^{1 + m}

Simplify

\frac{1}{m} [- \frac{1}{u}]_{1}^{1 + m} : \frac{[ - \frac{1}{u} ] _{1}^{1 + m}}{m}

= \frac{[ - \frac{1}{u} ] _{1}^{1 + m}}{m}

Compute the boundaries:

- \frac{1}{1 + m} + 1

= \frac{- \frac{1}{1 + m} + 1}{m}

Simplify

= \frac{1}{1 + m}

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What is the integral from 0 to 1 of 1/((mx+1)^2) ?
The integral from 0 to 1 of 1/((mx+1)^2) is 1/(1+m)