Integrals MCQs

∫1 dx means find a function whose derivative is 1.

The derivative of x is 1, so x is an antiderivative of the integrand.

Since indefinite integrals are defined up to an arbitrary constant, add C.

Hence ∫1 dx = x + C.

Recall the derivative of arctan x is 1/(1+x²).

Integrate both sides: ∫1/(1+x²)dx = arctan x + C.

Other listed antiderivatives differentiate to different functions (e.g., d/dx sin⁻¹x = 1/√(1−x²)).

Thus the correct antiderivative is tan⁻¹(x) + C.

Let u = x², then du = 2x dx, so the integral becomes ∫cos u du.

Integrating cos u gives sin u + C.

Replace u with x² to obtain sin(x²) + C, which matches the given answer.

Let u = x², then du = 2x dx, so the integral becomes ∫ du/(1+u²).

The antiderivative of 1/(1+u²) is arctan u.

Replace u by x² to get arctan(x²) + C, which matches the given answer. tan^-1(x^2) + C

∫log x dx set u=log x ⇒ du=dx/x, so write ∫log x·dx = ∫log x·x·(dx/x)=∫x·log x·(dx/x).

Integrate by parts: let A=log x (so dA=dx/x) and dB=x dx (so B=x²/2). Then ∫log x dx = x·log x−∫x·(dx/x)=x·log x−∫dx = x·log x−x + C.

Thus the antiderivative is x log x−x+C.

Multiply both sides by (x‑1)(x+2): 1 = A(x+2) + B(x‑1).

Set x = 1 to eliminate B: 1 = A(1+2) ⇒ A = 1/3.

Thus the coefficient A equals 1/3, matching the given answer.

∫₀¹ x² dx = [x³/3]₀¹

Evaluate the antiderivative at the limits: (1³/3) – (0³/3)

= 1/3 – 0 = 1/3

Hence the correct answer is 1/3.

Recall the derivative of arcsin x: d/dx[sin⁻¹x]=1/√(1−x²).

Since the integrand is exactly this derivative, the antiderivative is sin⁻¹x plus a constant.

Thus ∫1/√(1−x²)dx = sin⁻¹x + C.

sin⁻¹(x) + C.

Recall ∫dx/(a²+x²)= (1/a) tan⁻¹(x/a)+C.

Here a²=9 so a=3, giving ∫dx/(x²+9)= (1/3) tan⁻¹(x/3)+C.

Thus the correct choice is (1/3) tan⁻¹(x/3) + C.

Rewrite the integrand as cos x dx/(1+sin x).

Let u=1+sin x ⇒ du=cos x dx, so the integral becomes ∫du/u.

Integrating gives ln|u|+C, and substituting back u=1+sin x yields log|1+sin x|+C.

Use integration by parts: let u = x, dv = sin x dx → du = dx, v = –cos x.

∫x sin x dx = u·v – ∫v du = (x)(–cos x) – ∫(–cos x)dx = –x cos x + ∫cos x dx.

∫cos x dx = sin x, so the result is –x cos x + sin x + C, which matches the first option.

Write 1/[(x‑1)(x‑2)] as partial fractions: 1/[(x‑1)(x‑2)]=A/(x‑1)+B/(x‑2).

Solve A(x‑2)+B(x‑1)=1 → (A+B)x‑(2A+ B)=1, so A+B=0 and –(2A+ B)=1 → A=–1, B=1.

Thus ∫1/[(x‑1)(x‑2)]dx =∫[–1/(x‑1)+1/(x‑2)]dx = –ln|x‑1|+ln|x‑2|+C = ln|(x‑2)/(x‑1)|+C.

Rewrite integrand by polynomial division: (x^2+1)/(x-1)=x+1+2/(x-1).

Integrate termwise: ∫x dx = x^2/2, ∫1 dx = x, ∫2/(x-1) dx = 2 ln|x‑1|.

Combine the antiderivatives and add constant C.

Thus the integral equals x^2/2 + x + 2 log|x‑1| + C.

∫_{-1}^{1} x^3 dx = [x^4/4]_{-1}^{1}

Evaluate at the limits: (1^4/4) – ((-1)^4/4) = (1/4) – (1/4) = 0

Since x^3 is an odd function, the area on [-1,0] cancels that on [0,1].

Hence the integral equals 0.

Apply the power rule for integration: ∫x^k dx = x^{k+1}/(k+1) + C provided k ≠ –1.

Here the integrand is x^n, so k = n.

Since n ≠ –1, the antiderivative is x^{n+1}/(n+1) + C, which matches the given answer.

Recall that ∫a^x dx is of the form ∫e^{x\ln a}dx.

Write a^x = e^{(\ln a)x} and integrate: ∫e^{(\ln a)x}dx = e^{(\ln a)x}/(\ln a)+C.

Replace e^{(\ln a)x} by a^x to obtain a^x/(\ln a)+C.

Thus the integral equals a^x/log a + C.

Write denominator as D = x^2+x+1, then D' = 2x+1 which is exactly the numerator.

∫(2x+1)/D dx = ∫ D'/D dx.

Integral of D'/D is ln|D| + C.

Hence the result is log|x^2 + x + 1| + C.

Let u = sin x, then du = cos x dx.

The integral becomes ∫u³ du = u⁴/4 + C.

Replace u by sin x to get (sin⁴ x)/4 + C, which matches the given answer.

Use integration by parts: let u = log x, dv = x dx.

Then du = (1/x)dx and v = x²/2.

∫x log x dx = (x²/2)·log x – ∫(x²/2)·(1/x)dx = (x²/2)·log x – ∫x/2 dx.

∫x/2 dx = x²/4, so the result is (x²/2) log x – x²/4 + C.

(x²/2) log x - x²/4 + C

Factor denominator: x^2−1=(x−1)(x+1).

Write 1/(x^2−1)=A/(x−1)+B/(x+1) and solve A+B=0, A−B=1 ⇒ A=½, B=−½.

Integrate termwise: ∫½·dx/(x−1)−½·dx/(x+1)=½ ln|x−1|−½ ln|x+1|+C.

Combine logs: (1/2) ln|(x−1)/(x+1)|+C.