Problem to solve
Find the indefinite integral \[\int x^4\cdot\sqrt{x^2+4x+5} dx\]
Solution
STEP 1
Take the integral: integral x^4 sqrt(x^2 + 4 x + 5) dx
STEP 2
For the integrand x^4 sqrt(x^2 + 4 x + 5), complete the square: = integral x^4 sqrt((x + 2)^2 + 1) dx
STEP 3
For the integrand x^4 sqrt((x + 2)^2 + 1), substitute u = x + 2 and du = dx: = integral(u - 2)^4 sqrt(u^2 + 1) du
STEP 4
For the integrand (u - 2)^4 sqrt(u^2 + 1), substitute u = tan(s) and du = sec^2(s) ds. Then sqrt(u^2 + 1) = sqrt(tan^2(s) + 1) = sec(s) and s = tan^(-1)(u): = integral(tan(s) - 2)^4 sec^3(s) ds
STEP 5
Expanding the integrand (tan(s) - 2)^4 sec^3(s) gives 16 sec^3(s) + tan^4(s) sec^3(s) - 8 tan^3(s) sec^3(s) + 24 tan^2(s) sec^3(s) - 32 tan(s) sec^3(s): = integral(16 sec^3(s) + tan^4(s) sec^3(s) - 8 tan^3(s) sec^3(s) + 24 tan^2(s) sec^3(s) - 32 tan(s) sec^3(s)) ds
STEP 6
Integrate the sum term by term and factor out constants: = integral tan^4(s) sec^3(s) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds + 16 integral sec^3(s) ds
STEP 7
For the integrand tan^4(s) sec^3(s), use the trigonometric identity tan^2(s) = sec^2(s) - 1: = integral sec^3(s) (sec^2(s) - 1)^2 ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds + 16 integral sec^3(s) ds
STEP 8
Expanding the integrand sec^3(s) (sec^2(s) - 1)^2 gives sec^7(s) - 2 sec^5(s) + sec^3(s): = integral(sec^7(s) - 2 sec^5(s) + sec^3(s)) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds + 16 integral sec^3(s) ds
STEP 9
Integrate the sum term by term and factor out constants: = integral sec^7(s) ds - 2 integral sec^5(s) ds + 17 integral sec^3(s) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 10
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m - 1)(s))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(s) ds, where m = 7: = 1/6 tan(s) sec^5(s) - 7/6 integral sec^5(s) ds + 17 integral sec^3(s) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 11
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m - 1)(s))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(s) ds, where m = 5: = -7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 129/8 integral sec^3(s) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 12
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m - 1)(s))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(s) ds, where m = 3: = 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 129/16 integral sec(s) ds - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 13
The integral of sec(s) is log(tan(s) + sec(s)): = 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 8 integral tan^3(s) sec^3(s) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 14
For the integrand tan^3(s) sec^3(s), use the trigonometric identity tan^2(s) = sec^2(s) - 1: = 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 8 integral tan(s) sec^3(s) (sec^2(s) - 1) ds + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 15
For the integrand tan(s) sec^3(s) (sec^2(s) - 1), substitute p = sec(s) and dp = tan(s) sec(s) ds: = 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 8 integral p^2 (p^2 - 1) dp + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 16
Expanding the integrand p^2 (p^2 - 1) gives p^4 - p^2: = 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 8 integral(p^4 - p^2) dp + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 17
Integrate the sum term by term and factor out constants: = 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 8 integral p^4 dp + 8 integral p^2 dp + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 18
The integral of p^4 is p^5/5: = -(8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 8 integral p^2 dp + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 19
The integral of p^2 is p^3/3: = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 24 integral tan^2(s) sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 20
For the integrand tan^2(s) sec^3(s), write tan^2(s) as sec^2(s) - 1: = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 24 integral sec^3(s) (sec^2(s) - 1) ds - 32 integral tan(s) sec^3(s) ds
STEP 21
Expanding the integrand sec^3(s) (sec^2(s) - 1) gives sec^5(s) - sec^3(s): = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 24 integral(sec^5(s) - sec^3(s)) ds - 32 integral tan(s) sec^3(s) ds
STEP 22
Integrate the sum term by term and factor out constants: = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) - 7/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) + 24 integral sec^5(s) ds - 24 integral sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 23
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m - 1)(s))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(s) ds, where m = 5: = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 129/16 tan(s) sec(s) + 137/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 6 integral sec^3(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 24
Use the reduction formula, integral sec^m(s) ds = (sin(s) sec^(m - 1)(s))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(s) ds, where m = 3: = (8 p^3)/3 - (8 p^5)/5 + 129/16 log(tan(s) + sec(s)) + 81/16 tan(s) sec(s) + 137/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 3 integral sec(s) ds - 32 integral tan(s) sec^3(s) ds
STEP 25
The integral of sec(s) is log(tan(s) + sec(s)): = (8 p^3)/3 - (8 p^5)/5 + 81/16 log(tan(s) + sec(s)) + 81/16 tan(s) sec(s) + 137/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 32 integral tan(s) sec^3(s) ds
STEP 26
For the integrand tan(s) sec^3(s), substitute w = sec(s) and dw = tan(s) sec(s) ds: = (8 p^3)/3 - (8 p^5)/5 + 81/16 log(tan(s) + sec(s)) + 81/16 tan(s) sec(s) + 137/24 tan(s) sec^3(s) + 1/6 tan(s) sec^5(s) - 32 integral w^2 dw
STEP 27
The integral of w^2 is w^3/3: = -(8 p^5)/5 + (8 p^3)/3 + 1/6 tan(s) sec^5(s) + 137/24 tan(s) sec^3(s) + 81/16 tan(s) sec(s) + 81/16 log(tan(s) + sec(s)) - (32 w^3)/3 + constant
STEP 28
Substitute back for w = sec(s): = -(8 p^5)/5 + (8 p^3)/3 - (32 sec^3(s))/3 + 1/6 tan(s) sec^5(s) + 137/24 tan(s) sec^3(s) + 81/16 tan(s) sec(s) + 81/16 log(tan(s) + sec(s)) + constant
STEP 29
Substitute back for p = sec(s): = -(8 sec^5(s))/5 - 8 sec^3(s) + 1/6 tan(s) sec^5(s) + 137/24 tan(s) sec^3(s) + 81/16 tan(s) sec(s) + 81/16 log(tan(s) + sec(s)) + constant
STEP 30
Substitute back for s = tan^(-1)(u): = 1/6 tan(tan^(-1)(u)) sec(tan^(-1)(u))^5 - 8/5 sec(tan^(-1)(u))^5 + 137/24 tan(tan^(-1)(u)) sec(tan^(-1)(u))^3 - 8 sec(tan^(-1)(u))^3 + 81/16 tan(tan^(-1)(u)) sec(tan^(-1)(u)) + 81/16 log(tan(tan^(-1)(u)) + sec(tan^(-1)(u))) + constant
STEP 31
Simplify using sec(tan^(-1)(z)) = sqrt(z^2 + 1) and tan(tan^(-1)(z)) = z: = 1/6 u (u^2 + 1)^(5/2) - 8/5 (u^2 + 1)^(5/2) + 137/24 u (u^2 + 1)^(3/2) - 8 (u^2 + 1)^(3/2) + 81/16 u sqrt(u^2 + 1) + 81/16 log(sqrt(u^2 + 1) + u) + constant
STEP 32
Substitute back for u = x + 2: = 1/6 (x + 2) ((x + 2)^2 + 1)^(5/2) - 8/5 ((x + 2)^2 + 1)^(5/2) + 137/24 (x + 2) ((x + 2)^2 + 1)^(3/2) - 8 ((x + 2)^2 + 1)^(3/2) + 81/16 (x + 2) sqrt((x + 2)^2 + 1) + 81/16 log(x + sqrt((x + 2)^2 + 1) + 2) + constant
STEP 33
Factor the answer a different way: = 1/240 (sqrt(x^2 + 4 x + 5) (40 x^5 + 16 x^4 - 22 x^3 - 4 x^2 + 185 x - 1070) + 1215 log(x + sqrt((x + 2)^2 + 1) + 2)) + constant
STEP 34
An alternative form of the integral is: Answer: | | = 1/240 (sqrt(x^2 + 4 x + 5) (40 x^5 + 16 x^4 - 22 x^3 - 4 x^2 + 185 x - 1070) + 1215 sinh^(-1)(x + 2)) + constant
More problems
Number
001Find the indefinite integral \(\int\frac{x^2+2}{x^3+5x^2+2x-8} dx\)
002Find the indefinite integral \(\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\)
003Find the indefinite integral \(\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx\)
004Find the indefinite integral \(\int\frac{3x^4+4x^2+5x}{(x-1)(x^2+1)^2} dx\)
005Find the indefinite integral \(\int\frac{x^6+6x^4+8x^2-4}{(x^2+4)(x^2+3)} dx\)
006Find the indefinite integral \(\int(-4x^4+6x^2+8x-1)\cos(x) dx\)
007Find the indefinite integral \(\int(-7x^5+2x^4-2)\cdot3^xdx\)
008Find the indefinite integral \(\int\arcsin^4(x) dx\)
009Find the indefinite integral \(\int\sin(\ln(x)) dx\)
010Find the indefinite integral \(\int(12x^{15}-3x^7-4x^3)\cdot e^{2x^4} dx\)
011Find the indefinite integral \(\int\frac{\sqrt{(3x^2+4)^3}}{x^4} dx\)
012Find the indefinite integral \(\int\frac{x^2}{\sqrt{2x^2-5}} dx\)
013Find the indefinite integral \(\int\frac{x^6}{\sqrt{(x^2+4)^3}} dx\)
014Find the indefinite integral \(\int x^2\sqrt{4-3x^2} dx\)
015Find the indefinite integral \(\int x^4\cdot\sqrt{x^2+4x+5} dx\)
016Find the indefinite integral \(\int\frac{\sqrt{x+2}}{x+1} dx\)
017Find the indefinite integral \(\int\frac{\sqrt{x+2}-4}{5-\sqrt{x+2}} dx\)
018Find the indefinite integral \(\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx\)
019Find the indefinite integral \(\int\sqrt{\frac{x+3}{1-x}} dx\)
020Find the indefinite integral \(\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx\)
021Find the indefinite integral \(\int\frac{\sin(x)-3\cos(x)}{\cos(x)-2} dx\)
022Find the indefinite integral \(\int\frac{3\sec(x)-\cot(x)}{1-\csc(x)} dx\)
023Find the indefinite integral \(\int\frac{\tan^2(x)+2}{\tan(x)+2} dx\)
024Find the indefinite integral \(\int\frac{3\sin^4(x)-2\cos^2(x)}{1+\sin^2(x)} dx\)
025Find the indefinite integral \(\int\tan^4(x)\sec^5(x) dx\)
Back
Edited by tododemate® - 2021