Visual SolutionPYQ 2024 · Jan Shift 1Standard

Question

The integral

\int \frac{(x^8 - x^2) \, dx}{(x^{12} + 3x^6 + 1) \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)}

equal to:

(A)

\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1/3} + C

(B)

\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1/2} + C

(C)

\log_e\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right| + C

(D)

\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{3} + C

Solution Path

Sub

t = \tan^{-1}(x^3+x^{-3})

reduces to

\frac{1}{3}\int dt/t

. Answer:

\ln|\tan^{-1}(x^3+x^{-3})|^{1/3}+C

01Question Setup

1/4

Evaluate

\int \frac{(x^8 - x^2)\,dx}{(x^{12}+3x^6+1)\,\tan^{-1}(x^3+\frac{1}{x^3})}

I = \;?

02Substitution

2/4

Let

t = \tan^{-1}(x^3 + x^{-3})

. Compute

dt

and simplify the denominator.

t = \tan^{-1}(x^3 + x^{-3})

03SimplifyKEY INSIGHT

3/4

Numerator

x^8 - x^2 = x^2(x^6 - 1)

matches

\frac{1}{3}\,dt

. Integral becomes

\frac{1}{3}\int \frac{dt}{t} = \ln|t|^{1/3} + C

\frac{1}{3}\ln|t| + C

04Final Answer

4/4

I = \ln\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1/3} + C

\boxed{\ln|\tan^{-1}(\cdots)|^{1/3} + C}

Concepts from this question1 concepts unlocked

★

EASY

Replace a composite expression with a single variable to simplify the integrand

\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad t = g(x)

Transforms complex integrands into standard forms that can be integrated directly

Trigonometric integralsRational functionsPower rule integrals

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