What is (1+1/(1+t^2))(1+t^2)^2 ?

The solution to (1+1/(1+t^2))(1+t^2)^2 is (t^2+2)(1+t^2)

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simplify (1+1/(1+t^2))(1+t^2)^2

simplify

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

(t^{2} + 2) (1 + t^{2})

Solution steps

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

Join

1 + \frac{1}{1 + t ^{2}} : \frac{t ^{2} + 2}{1 + t ^{2}}

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

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(1 + t^{2})^{2} = (1 + t^{2}) (1 + t^{2})

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

Cancel the common factor:

1 + t^{2}

= (t^{2} + 2) (1 + t^{2})

\frac{ln ( \frac{( 1 + x + h )}{1 + x} )}{h}

\frac{1459}{( 1 + x ) ^{10}}

(x^{2} + 1 - x^{2} + 3 x + 4)^{3}

\frac{d}{d y} ((z) (3 x^{2} y^{2} + 2 z^{2} x y - 2 z x^{3} + 4 z y^{3} - 4))

(\frac{- 36}{4} bi g)^{2} (4 - 36) 2

What is (1+1/(1+t^2))(1+t^2)^2 ?
The solution to (1+1/(1+t^2))(1+t^2)^2 is (t^2+2)(1+t^2)