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Problem to solve

Find the indefinite integral \[\int\frac{\tan^2[x]+2}{\tan[x]+2} dx\]

Solution

STEP 1

Take the integral:
integral(tan^2(x) + 2)/(tan(x) + 2) dx

STEP 2

Multiply numerator and denominator of (tan^2(x) + 2)/(tan(x) + 2) by sec^2(x):
= integral(2 sec^2(x) + tan^2(x) sec^2(x))/(2 sec^2(x) + tan(x) sec^2(x)) dx

STEP 3

Prepare to substitute u = tan(x). Rewrite (2 sec^2(x) + tan^2(x) sec^2(x))/(2 sec^2(x) + tan(x) sec^2(x)) using sec^2(x) = tan^2(x) + 1:
= integral((tan^2(x) + 2) sec^2(x))/(tan^3(x) + 2 tan^2(x) + tan(x) + 2) dx

STEP 4

For the integrand ((tan^2(x) + 2) sec^2(x))/(tan^3(x) + 2 tan^2(x) + tan(x) + 2), substitute u = tan(x) and du = sec^2(x) dx:
= integral(u^2 + 2)/(u^3 + 2 u^2 + u + 2) du

STEP 5

$For the integrand (u^2 + 2)/(u^3 + 2 u^2 + u + 2), use partial fractions: = integral(2 - u)/(5 (u^2 + 1)) + 6/(5 (u + 2)) du$

STEP 6

Integrate the sum term by term and factor out constants:
= 1/5 integral(2 - u)/(u^2 + 1) du + 6/5 integral1/(u + 2) du

STEP 7

Expanding the integrand (2 - u)/(u^2 + 1) gives 2/(u^2 + 1) - u/(u^2 + 1):
= 1/5 integral(2/(u^2 + 1) - u/(u^2 + 1)) du + 6/5 integral1/(u + 2) du

STEP 8

Integrate the sum term by term and factor out constants:
= -1/5 integral u/(u^2 + 1) du + 2/5 integral1/(u^2 + 1) du + 6/5 integral1/(u + 2) du

STEP 9

For the integrand u/(u^2 + 1), substitute s = u^2 + 1 and ds = 2 u du:
= -1/10 integral1/s ds + 2/5 integral1/(u^2 + 1) du + 6/5 integral1/(u + 2) du

STEP 10

The integral of 1/s is log(s):
= -log(s)/10 + 2/5 integral1/(u^2 + 1) du + 6/5 integral1/(u + 2) du

STEP 11

The integral of 1/(u^2 + 1) is tan^(-1)(u):
= 2/5 tan^(-1)(u) - log(s)/10 + 6/5 integral1/(u + 2) du

STEP 12

For the integrand 1/(u + 2), substitute p = u + 2 and dp = du:
= 2/5 tan^(-1)(u) - log(s)/10 + 6/5 integral1/p dp

STEP 13

The integral of 1/p is log(p):
= (6 log(p))/5 - log(s)/10 + 2/5 tan^(-1)(u) + constant

STEP 14

Substitute back for p = u + 2:
= -log(s)/10 + 6/5 log(u + 2) + 2/5 tan^(-1)(u) + constant

STEP 15

Substitute back for s = u^2 + 1:
= -1/10 log(u^2 + 1) + 6/5 log(u + 2) + 2/5 tan^(-1)(u) + constant

STEP 16

Substitute back for u = tan(x):
= (2 x)/5 + 6/5 log(tan(x) + 2) - 1/10 log(sec^2(x)) + constant

STEP 17

Factor the answer a different way:
= 1/10 (4 x + 12 log(tan(x) + 2) - log(sec^2(x))) + constant

STEP 18

Which is equivalent for restricted x values to:
Answer: |
| = 1/5 (2 x - 5 log(cos(x)) + 6 log(sin(x) + 2 cos(x))) + constant

More problems

Number
001	Find the indefinite integral $\int\frac{x^2+2}{x^3+5x^2+2x-8} dx$
002	Find the indefinite integral $\int\frac{4x^4-2}{x^3+4x^2+x-6} dx$
003	Find the indefinite integral $\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx$
004	Find the indefinite integral $\int\frac{3x^4+4x^2+5x}{(x-1)(x^2+1)^2} dx$
005	Find the indefinite integral $\int\frac{x^6+6x^4+8x^2-4}{(x^2+4)(x^2+3)} dx$
006	Find the indefinite integral $\int(-4x^4+6x^2+8x-1)\cos(x) dx$
007	Find the indefinite integral $\int(-7x^5+2x^4-2)\cdot3^xdx$
008	Find the indefinite integral $\int\arcsin^4(x) dx$
009	Find the indefinite integral $\int\sin(\ln(x)) dx$
010	Find the indefinite integral $\int(12x^{15}-3x^7-4x^3)\cdot e^{2x^4} dx$
011	Find the indefinite integral $\int\frac{\sqrt{(3x^2+4)^3}}{x^4} dx$
012	Find the indefinite integral $\int\frac{x^2}{\sqrt{2x^2-5}} dx$
013	Find the indefinite integral $\int\frac{x^6}{\sqrt{(x^2+4)^3}} dx$
014	Find the indefinite integral $\int x^2\sqrt{4-3x^2} dx$
015	Find the indefinite integral $\int x^4\cdot\sqrt{x^2+4x+5} dx$
016	Find the indefinite integral $\int\frac{\sqrt{x+2}}{x+1} dx$
017	Find the indefinite integral $\int\frac{\sqrt{x+2}-4}{5-\sqrt{x+2}} dx$
018	Find the indefinite integral $\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx$
019	Find the indefinite integral $\int\sqrt{\frac{x+3}{1-x}} dx$
020	Find the indefinite integral $\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx$
021	Find the indefinite integral $\int\frac{\sin(x)-3\cos(x)}{\cos(x)-2} dx$
022	Find the indefinite integral $\int\frac{3\sec(x)-\cot(x)}{1-\csc(x)} dx$
023	Find the indefinite integral $\int\frac{\tan^2(x)+2}{\tan(x)+2} dx$
024	Find the indefinite integral $\int\frac{3\sin^4(x)-2\cos^2(x)}{1+\sin^2(x)} dx$
025	Find the indefinite integral $\int\tan^4(x)\sec^5(x) dx$

Back

Number
001	Find the indefinite integral \(\int\frac{x^2+2}{x^3+5x^2+2x-8} dx\)
002	Find the indefinite integral \(\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\)
003	Find the indefinite integral \(\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx\)
004	Find the indefinite integral \(\int\frac{3x^4+4x^2+5x}{(x-1)(x^2+1)^2} dx\)
005	Find the indefinite integral \(\int\frac{x^6+6x^4+8x^2-4}{(x^2+4)(x^2+3)} dx\)
006	Find the indefinite integral \(\int(-4x^4+6x^2+8x-1)\cos(x) dx\)
007	Find the indefinite integral \(\int(-7x^5+2x^4-2)\cdot3^xdx\)
008	Find the indefinite integral \(\int\arcsin^4(x) dx\)
009	Find the indefinite integral \(\int\sin(\ln(x)) dx\)
010	Find the indefinite integral \(\int(12x^{15}-3x^7-4x^3)\cdot e^{2x^4} dx\)
011	Find the indefinite integral \(\int\frac{\sqrt{(3x^2+4)^3}}{x^4} dx\)
012	Find the indefinite integral \(\int\frac{x^2}{\sqrt{2x^2-5}} dx\)
013	Find the indefinite integral \(\int\frac{x^6}{\sqrt{(x^2+4)^3}} dx\)
014	Find the indefinite integral \(\int x^2\sqrt{4-3x^2} dx\)
015	Find the indefinite integral \(\int x^4\cdot\sqrt{x^2+4x+5} dx\)
016	Find the indefinite integral \(\int\frac{\sqrt{x+2}}{x+1} dx\)
017	Find the indefinite integral \(\int\frac{\sqrt{x+2}-4}{5-\sqrt{x+2}} dx\)
018	Find the indefinite integral \(\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx\)
019	Find the indefinite integral \(\int\sqrt{\frac{x+3}{1-x}} dx\)
020	Find the indefinite integral \(\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx\)
021	Find the indefinite integral \(\int\frac{\sin(x)-3\cos(x)}{\cos(x)-2} dx\)
022	Find the indefinite integral \(\int\frac{3\sec(x)-\cot(x)}{1-\csc(x)} dx\)
023	Find the indefinite integral \(\int\frac{\tan^2(x)+2}{\tan(x)+2} dx\)
024	Find the indefinite integral \(\int\frac{3\sin^4(x)-2\cos^2(x)}{1+\sin^2(x)} dx\)
025	Find the indefinite integral \(\int\tan^4(x)\sec^5(x) dx\)