30.03.2023 - 16:44

Solve this calculus problem. d t 4 ( tan 1 t ) ( 1 + t 2 ) A. I n 4 tan 1 t + C B. 4 cot 1 t + C C. 1 4 I n tan 1 t + C D. 1 4 ( tan 1 t ) 2 + C

Solve this calculus problem.

{eq}\int \frac{dt}{4(tan ^{-1}t)(1+t^{2})} {/eq}

A. {eq}In \left | 4tan ^{-1}t \right | + C {/eq}

B. {eq}4 cot ^{-1}t + C {/eq}

C.{eq}\frac{1}{4} In \left | tan ^{-1}t \right |+ C {/eq}

D. {eq}\frac{1}{4(tan ^{-1}t)^{2}} + C {/eq}

Answers (1)
  • dj-prado-dk
    April 1, 2023 в 23:55

    The given calculus problem can be solved using the substitution method. Let u = an^-1t, then du/dt = -an^-2 and dt = -(1/a)u^2du. Substituting these values in the integral, we get:

    ∫dt/[4(an^-1t)(1+t^2)] = -1/(4a) ∫du/[u(1+(u/an)^2)]

    Now, let v = u/an, then dv/du = 1/an and du = anv dv. Substituting these values, we get:

    -1/(4a) ∫dv/[v(1+v^2)] = -1/(4a) * [1/2 ln|1+v^2|] + C

    Substituting back the value of v, we get:

    -1/(4a) * [1/2 ln|1+(u/an)^2|] + C = -1/(8a) ln|an^-1t|^2 + C

    Simplifying this expression, we get:

    -1/(8a) ln(an^-1t)^2 + C = -1/4 ln|an^-1t| + C = ln|an^-1t|^(-1/4) + C

    Therefore, the answer is C. {eq}\frac{1}{4} ln \left|an^{-1}t \right| + C {/eq}.

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